Transmission Lines and Maxwell's Equations

1 Transmission Lines and Maxwell's Equations Phil Lucht Rimrock Digital Technology, Salt Lake City, Utah 84103 last update: Aug 2, 2014 Maple code is available upon request. Comments and errata are welcome. The material in this document is copyrighted by the author. The graphics look ratty in Windows Adobe PDF viewers wh en not scaled up, but look just fine in this excellent freeware viewer: http://www.tracker-software.com/pdf-xchange-products-co mparison-chart . The table of contents has live links. Most PDF view ers provide these links as bookmarks on the left. Overview an d Summary ........................................................................................................... .............. 7 Chapter Summaries.............................................................................................................. ................. 8 Appendix Summaries............................................................................................................. ............. 10 Symbols used in this document........................................................................................................... 12 Chapter 1: Basi c Equations..................................................................................................... ............. 17 1.1 Maxwell's Equations in a Conducting Dielectric Medium ........................................................... 17 (a) Notes on Maxwell's Equations ............................................................................................... ...17 (b) Integral Forms of Maxwel l's Equations and Continuity............................................................ 22 (c) Rules for behavior of fiel ds and potentials at a boundary ......................................................... 24 1.2 The Field Wave Equations................................................................................................... ......... 29 1.3 The Potential Wave Equations............................................................................................... ....... 30 (a) The Potential Wave Equati ons in the Lorenz gauge.................................................................. 30 (b) Special Relativity Note.............................................................................................................. 32 (c) The Potential Wave Equations in th e King and Lorenz Gauges with Conductors .................... 34 1.4 Retarded Solutions in th e Lorenz gauge: Propagators ................................................................. 43 1.5 The Wave Equations in the Frequency Domain ........................................................................... 46 (a) The Transformed Wave Equations............................................................................................ 46 (b) The Helmholtz Integrals in the King Gauge.............................................................................. 48 (c) King's leading factor (1/4πξ ) and the final Helmholtz Integrals................................................ 49 (d) Frequency domain wave equations for fields and potentials in the Lorenz Gauge................... 57 (e) Self Consistency of He lmholtz Integral Solutions..................................................................... 58 1.6 Reinterpretation of all equations in terms of complex functions .................................................. 60 (a) Complex Functions.......................................................................................................... .......... 60 (b) Monochrome time ............................................................................................................ ......... 61 (c) Why complex fields: The Fourier Transform........................................................................... 62 (d) Monochrome E and B fields.................................................................................................. ....63 (e) A Pitfall to Avoid ......................................................................................................... ............. 64 (f) Overloaded Notation and Maxwell's Equations in ω space ....................................................... 64 Chapter 2: The Round Wi re and the Ski n Effect............................................................................... 65 2.1 The Implicit Wave Context, Helmho ltz Equations and the Skin Effect ....................................... 65 2.2 Derivation of E(r), B(r) and J(r) for a round wire......................................................................... 68 2 2.3 A study of the solution of a round wire .................................................................................... ....75 (a) Kelvin Functions ........................................................................................................... ............ 75 (b) Plots of |E(r)/E(a)| for various δ values .................................................................................... 76 (c) Review of the round wire solution .......................................................................................... ..80 (d) Plots of the round wire solution for Belden 8281 at 5 MHz...................................................... 82 2.4 The Surface Impedance Z s(ω) of a Round Wire........................................................................... 84 (a) Expressions for Surface Impedance.......................................................................................... .85 (b) Low frequency limit of Z s(ω).................................................................................................... 86 (c) High frequency limit of Z s(ω) ................................................................................................... 87 (d) Plots of Z s(ω) versus skin depth δ............................................................................................. 88 2.5 Surface Impedance for a Transmission Line................................................................................. 91 Chapter 3: Transmission Line Preliminaries ..................................................................................... 94 3.1 Why is there no free charge in side a conductor or a dielectric? ................................................... 94 3.2 How thick is the surface charge layer on a conductor?................................................................. 96 3.3 How does loss tangent aff ect dielectric conductivity?.................................................................. 97 3.4 Size of E fields in conductor and dielectri c; conservation of total current at a boundary............. 99 3.5 The TEM mode fields and currents for an ideal transmission line ............................................. 101 3.6 The TEM mode fields and currents for a practical transmission line......................................... 104 3.7 The general shape of fields, charges, and currents on a transmission line.................................. 106 (a) Eθ at a conductor surface vanishes .......................................................................................... 106 (b) The transverse vector pot ential components are small............................................................ 107 (c) The scalar potential φ on a conductor surface ......................................................................... 107 (d) B and A z on a conductor surface............................................................................................. 111 A Counter Example and Comments on the Low Frequency Regime ........................................... 113 (e) Observations about the E and B field lines in a transmission line dielectric........................... 116 (f) Drawings of the fields..................................................................................................... ......... 117 (g) More on the field and current structure ................................................................................... 120 (h) Estimate of the ratio J r/Jz........................................................................................................ 121 3.8 Review of Transmission Line Preliminaries............................................................................... 123 Chapter 4: Transmission Line Equations......................................................................................... 126 4.1 Computation of potential φ due to one conductor of a transmission line ................................... 126 4.2 Computation of potential φ due to both conductors of a transmission line ................................ 127 4.3 The Transmission Line Limit...................................................................................................... 128 4.4 General Calculation of V(z)................................................................................................ ........ 131 4.5 Example: Transmission line with widely-spaced round wires of unequal diameters ................ 135 Power Transmission Lines (als o Telephone and Telegraph) ........................................................ 140 4.6 Example: A coaxial cable .................................................................................................. ........ 141 4.7 Computation of A z due to one conductor of a transmission line ................................................ 144 Comments regarding μ .................................................................................................................. 146 4.8 Computation of potential A z due to both conductors of a transmission line .............................. 148 4.9 Transmission Line Limit Revisited.......................................................................................... ...148 4.10 General Calculation of W(z)............................................................................................... ......148 4.11 The Classical Transmission Line Equations ............................................................................. 151 (a) Initial Processing ..................................................................................................................... 152 (b) Averaging Repair and the Transmission Line Equations ........................................................ 153 3 (c) Digression on the meaning of C'............................................................................................ ..158 (d) An example of K = K L............................................................................................................. 159 (e) Summary of Results......................................................................................................... ........ 162 (f) Time domain equations (telegraph equations) ......................................................................... 164 4.12 Modifications to account for μd ≠ μ1 ≠ μ2................................................................................. 165 Chapter 5: The Transverse Problem.............................................................................................. ...170 5.1 Separation of φ........................................................................................................................... 170 5.2 Separation of A z......................................................................................................................... 172 5.3 Development of the Transverse Problem.................................................................................... 174 (a) kφ = kA and the transverse equations........................................................................................ 174 (b) The scaling boundary condition on φt(x)................................................................................ 176 (c) Energy Conservation in a Transmission Line.......................................................................... 178 5.4 The Low-Loss Approximation................................................................................................. ...179 (a) Transverse Equations for a Low-Loss transmission line......................................................... 179 (b) The scaling boundary cond ition (5.3.13) revisited.................................................................. 182 5.5 The Capacitor Problem ...................................................................................................... ......... 183 5.6 What happens if lo w-loss is not assumed?.................................................................................. 189 Chapter 6: Two Cyli ndrical Conductors .......................................................................................... 192 6.1 A candidate transverse potential φt............................................................................................ 192 6.2 Ancient Greece circa 230 BC................................................................................................ ......192 6.3 Back to the Future: Calculation of K ...................................................................................... ...195 6.4 Summary of Line Parameter Results .......................................................................................... 202 6.5. The Proximity Effect for a Transm ission Line made of Two Round Wires .............................. 203 (a) The surface charge de nsity and its moments ........................................................................... 203 (b) The Proximity Effect....................................................................................................... ........ 206 (c) Plots of the Proximity and Skin Effects................................................................................... 207 (d) The relationship between J z(a,θ) and n(θ) obtained from div E = 0........................................ 212 (e) The Proximity Effect At Low Frequencies.............................................................................. 214 (f) Active perimeter p and Z s for a two-cylinder transmission line.............................................. 215 (g) The Proximity Effect For Currents in the Same Direction...................................................... 218 Appendix A: Gauge Invariance ................................................................................................... ......219 A.0 The Poisson Equation and its Solution...................................................................................... .219 A.1 Existence of A such that B = curl A and div A = 0.................................................................... 222 A.2 Existence of A' such that B = curl A' and div A' = f . ................................................................ 224 A.3 Existence of φ such that E = -grad φ.......................................................................................... 224 A.4 Existence of A' and φ' such that B = curl A', E = - grad φ' - ∂tA', and div A' = f...................... 225 A.5 Gauge Invariance ........................................................................................................... ............ 226 A.6 The Lorenz Gauge and QED................................................................................................... ...227 A.7 Finding the gauge function Λ for the Lorentz Gauge: time-domain propagators ..................... 229 Appendix B: Magnetization Surface Currents on a Conductor ..................................................... 232 B.1 Relationship between surface current K and the field H at a conductor boundary .................... 232 B.2 Calculation of H from the current J in a conductor.................................................................... 236 (a) An expression for H in terms of J.......................................................................................... ..236 (b) An alternative derivation using the vector potential A z.......................................................... 237 4 (c) Boundary conditions........................................................................................................ ........ 238 (d) The Biot-Savart Law in 3D and 2D......................................................................................... 239 B.3 General Method for computing the surface current J m on a wire................................................ 240 B.4 Surface current on a round wire with uniform J......................................................................... 241 B.5 Computing H θ for a round wire using th e General Method of B.3 ............................................ 242 B.6 Modification of King's Helmholtz integral solution when μ1 ≠ μ2............................................ 245 (a) General Discussion......................................................................................................... ......... 245 (b) Statement and Proof of the J m Lemma..................................................................................... 247 (c) Statement and Proof of the J m Theorem................................................................................... 253 B.7 Application of the J m Lemma to a round wire with uniform J z................................................. 257 (a) The A z(c) term......................................................................................................................... 258 (b) The A z(m) term ........................................................................................................................ 260 (c) Adding the two terms and checking boundary conditions....................................................... 260 (d) Plots of A z and B θ and H θ....................................................................................................... 262 Appendix C: DC Properties of a Wire ............................................................................................ ..264 C.1 The DC resistance of a wire ................................................................................................ ....... 264 C.2 The DC surface im pedance of a wire ......................................................................................... 264 C.3 The DC internal and external inductance of a round wire.......................................................... 265 C.4 The DC internal inductance of a wire of rectangular cross section............................................ 268 C.5 The DC internal inducta nce of a thin flat wire........................................................................... 276 C.6 The DC internal inductance of a hollow round wire .................................................................. 281 Appendix D: The General E and B Fields Inside an Infinite Straight Round Wire ..................... 286 D.1 Partial Wave Expansion ..................................................................................................... ........ 287 (a) The General Method......................................................................................................... ....... 287 (b) Partial Wave Expansions.................................................................................................... .....289 (c) The Vector Laplacian in Cylindrical Coordinates ................................................................... 290 (d) The three Helmholtz equations and div E = 0 (in partial waves) ............................................ 291 D.2 Solutions for E z,Er and E θ......................................................................................................... 295 (a) The E z Solution ...................................................................................................................... .295 (b) The E r Solution ...................................................................................................................... .296 (c) The E θ Solution ...................................................................................................................... .298 (d) The Charge Pumping Boundary Condition ............................................................................. 301 (e) Application of the Boundary Conditions................................................................................. 302 Second summary of th e E field solutions...................................................................................... 306 D.3 What about the E θ Helmholtz Equation ? .................................................................................. 308 D.4 Computation of the B fields in the round wire........................................................................... 309 D.5 Verification that the E and B fi elds satisfy the Maxwell equations ........................................... 312 D.6 The exact E and B fields for the m=0 partial wave................................................................... 314 D.7 What about the E fields outside the round wire? ....................................................................... 315 D.8 About the boundary condition E θ(a,m) = 0................................................................................ 317 (a) The Quasi-Static Argument .................................................................................................. ...319 (b) An Ansatz Argument......................................................................................................... ......320 D.9 About the boundary condition E r(a,θ) = (jω/σ) n(θ) ................................................................. 322 (a) The notion of De bye Surface Currents .................................................................................... 322 (b) The role of Debye Surface Currents in the boundary condition.............................................. 323 5 (c) Where does surface charge n( θ) come fro m? .......................................................................... 325 (d) Modifications for a Conducting Di electric.............................................................................. 328 D.10 High frequency limit of the round wire E fields ..................................................................... 331 (a) Symmetry of f m, gm and hm and expansions for E i(r,θ)............................................................. 331 (b) High frequency evaluation of f m, gm and hm and the E fields.................................................... 332 D.11 Low frequency limit of the round wire E fields....................................................................... 337 (a) A High Level Review of Appendix D and its Accuracy ......................................................... 337 (b) Low frequency values for β'.................................................................................................... 338 (c) Low frequency evaluation of f m, gm and hm.............................................................................. 339 (d) Low frequenc y E fields ..................................................................................................... ......340 Appendix E: How Thick is Surface Charge on a Metal Conductor? ............................................. 343 Appendix F: Waveguides.................................................................................................................... 347 F.1 Discussion................................................................................................................. .................. 347 F.2 The TE waveguide modes for a parallel-plate transmission line................................................ 347 Appendix G: The DC vector potential of a round wire carrying a uniform current.................... 352 G.1 Setup and Assumptions...................................................................................................... ........ 352 G.2 Direct solution for A z(r) from the differential equation............................................................. 353 G.3 Instant solution for A using Ampere's Law and computation of J m............................................ 356 The Magnetization Current ...................................................................................................... .....356 G.4 Solution for A z using the 2D Helmholtz Integral ...................................................................... 357 G.5 Comments on the low frequency solution for A z....................................................................... 361 Appendix H: Poisson and Helmholtz Propagators in 3D ................................................................ 363 Appendix I: Poisson and Helmholtz Propagators in 2D.................................................................. 368 Appendix J: The 3D →2D Propagator Transition............................................................................ 376 Appendix K: The Network Model: Compar ison of Network and Maxwell Views ...................... 382 (a) The Network Model .......................................................................................................... .......... 382 (b) Network Model Characteristic Impedance.................................................................................. 383 (c) Network Model Transm ission Line Equations............................................................................ 385 (d) Network Model Parameters obt ained from Maxwell's Equations............................................... 386 (e) Low frequency case (no skin effect)........................................................................................ ....387 (f) High frequency case for round cond uctor (strong skin effect) .................................................... 388 Appendix L: Point and Line Charges in Dielectrics ........................................................................ 390 L.1 The potential of a point charge inside a thick dielectric spherical shell. .................................... 390 L.2 Limits of the Previous Problem............................................................................................. .....395 (a) Point charge in a spheri cal cavity in a dielectric ..................................................................... 395 (b) Point charge embedded in a dielectric sphere ......................................................................... 396 (c) Point charge embedded in an infinite dielectric medium ........................................................ 397 L.3 The potential of a line charge inside a thick dielectric cylindrical shell..................................... 398 L.4 Limits of the Previous Problem............................................................................................. .....401 (a) Line charge in an infinite cylindrical hole in a dielectric ........................................................ 401 (b) Line charge embedded in an infinite dielectric cylinder ......................................................... 402 (c) Line charge embedded in an infinite dielectric medium ......................................................... 402 Appendix M: Why the transverse vector potential A t is small for a transmission line................ 404 6 Appendix N: Drude, Magnetic Ohm's Law, Regular Hall Effect, Radial Hall Effect .................. 410 N.1 The Drude Model of Conduction .............................................................................................. .410 N.2 A Theory of the Hall Effect ....................................................................................................... 412 N.3 The Cyclotron Frequency.................................................................................................... ....... 416 N.4 Steady-state Electron Motion with E and B fields: Magnetic Ohm's Law................................ 417 N.5 Theory of the Hall Effect Revisited ....................................................................................... ...419 N.6 Theory of the Hall Effect with Multiple Carrier Types ............................................................ 420 N.7 The Radial Hall Effect in a Round Wire................................................................................... 424 N.8 Magnetic Ohm's La w for Arbitrary B ....................................................................................... 431 Appendix O: How to plot 2D magnetic field lines............................................................................ 434 (a) Statement of the Problem ................................................................................................... ......... 434 (b) The Brute Force Method ..................................................................................................... ........ 434 (c) The ODE Method ............................................................................................................. ........... 435 Example: Magnetic field lines for a two-cylinder transmission line ............................................... 436 (d) The Analytic Method ........................................................................................................ .......... 441 Appendix P: Eddy Currents and the Proximity Effect.................................................................... 443 P.1 Eddy Current Analysis...................................................................................................... .......... 444 P.2 Eddy currents in a thin round plate in a uniform B field ............................................................ 447 P.3 Eddy currents in a thin round plate in a non-uniform B field..................................................... 451 (a) The stream function method in Cartesian Coordinates............................................................ 452 (b) The stream function method in Cylindrical Coordinates ........................................................ 453 (c) Using the stream function method to solve the plate problem ................................................ 454 P.4 Self-induced eddy currents in a round wire ................................................................................ 458 P.5 Eddy currents induced in a quiet round-wire by an external B field .......................................... 462 P.6 Eddy currents induced in an current- carrying wire by an external B field ................................. 465 P.7 Summary of Round Wire Examples ........................................................................................... 466 P.8 Eddy currents in Transmission Lines: The Proximity Effect...................................................... 466 P.9 Quantitative Evaluation of Eddy Currents and The Proximity Effect ........................................ 469 P.10 Influence of Proximity and Skin Effects on Wire Resistance................................................... 470 Appendix Q: Properties of the functions k( ω) and Z 0(ω) ............................................................... 472 (a) Properties of k( ω) ....................................................................................................................... 472 (b) Properties of Z 0(ω) ..................................................................................................................... 476 Appendix R: Belden 8281 Coaxial Cable, a Case Study.................................................................. 482 (a) Geometry of the cable...................................................................................................... ............ 482 (b) Capac itance C.............................................................................................................. ................ 483 (c) Conductance G .............................................................................................................. .............. 484 (d) External inductance L e................................................................................................................ 485 (e) Total DC Inductance........................................................................................................ ............ 485 (f) High Frequency Inductance and Resistance ................................................................................ 486 (g) The Tinning Correction..................................................................................................... .......... 487 (h) Characteristic Impedance ................................................................................................... ......... 490 (i) Phase Velocity and Attenuation ............................................................................................. ......493 References............................................................................................................................................ 498 Overview and Summary 7 Overview and Summary This m onograph uses the Maxwell and associated poten tial equations to determine the behavior of infinitely-long, straight transmission lines. The presentation is loosely based on R.W.P. King's book Transmission-Line Theory . No attempt is made to address non-stra ight geometries, bends, stubs and many other practical applications described by King. Th ere is no discussion of discontinuities, reflections, standing wave ratios, Smith charts, or any of the tr aditional topics associated with transmission lines (see, for example, Pozar 2012). The emphasis is more on how one derives the transmission line parameters R,L,G,C directly from electromagnetic theory, and what approximations are made in doing so. A key requirement is that the wavelength of a transmission line wave be significantly larger than the line's transverse dimensions, something we refer to as th e "transmission line limit". Although the discussion generally concerns transmission lines with two conductors, comments here and there show how the conclusions can be extended to transmission lines with more than two conductors. Unlike waveguides, low-loss TEM transmission lines are most easily analyzed using potentials rather than fields due to the nature of the boundary conditions. This then brings up the can of worms known as "the gauge condition". We s how how a variant of the Lorenz gauge which we call "the King gauge" (since King uses it) serves to clarify the meaning of the Helm holtz integrals for the scalar and vector potentials over the surface and interior of the transmission line conductors. This subject is somewhat glossed over in King's highly compressed theoretical summary, and we could not find clarification in his many other books on the subject. By the way, most books on "tra nsmission lines" are concerned with the practical aspects of electrical power distribution and King's book is somewhat of a rarity, though there are other good books on the subject. It is true that a waveguide is in fact a transmission line, but we use the term "transmission line" to imply the TEM mode of transmission. An ancillary topic receiving much attention in this document is the description of the fields, potentials and currents inside a transmission line conductor operating at angular frequency ω. Mainly the discussion concerns round wires. A uniform round wire seems a simple physical object, yet the analysis is quite complicated and involves the so-called Kelvin functions . The skin effect and surface impedance of such a wire are considered in detail, and then late r the proximity effect enters the picture. There are very few "it can be shown" phrases in this document. Almost everything is derived in detail and the results verified against external sources. Simple examples are always presented and calculations for these examples are fully displayed, perhaps to a level of detail the reader will find annoying. Our view is that a piece of theory is useless if one cannot apply it to a simple case and get a reasonable result. The reader is assumed to have some knowledge of ordinary and partial differential equations and associated calculus. Green's Functions (which we call propagators) appear frequently, since these are useful in solving differential equations, and details ar e provided for readers not familiar with this subject. In particular, our first major waypoint is the deriva tion of the transmission line potentials in the form of King's Helmholtz integrals as shown in box (1.5.23) . The propagators in these integrals are the 3D Helmholtz free-space fundamental solutions e -jβR/R. This subject is fully laid out for the interested reader in Appendices H and I for the 3D and 2D Helmholtz partial differential equatio ns which are the frequency domain Fourier transforms of the more familiar 3D and 2D wave equations. Overview and Summary 8 The document consists of six Chapters which are followed at the end by Appendices A through R. The latter deal with issues thought too detailed or perhaps too peripheral to the main topic to appear in the main text, but which we nevertheless felt were wort h including. Many of the appendices are stand-alone monographs in their own right, addressing some related topic (eddy currents, gauge invariance, field line plotting methods, Hall effects, network model, fields in side a round wire, etc. ) The final section contains a list of References. Maple is used as needed to comput e analytic integrals, solve equations, do unpleasant algebra, and make graphs. The reader need not be a Maple expert to read and understand th e presented Maple code. To reduce clutter, derivatives that would normally be written ∂f ∂x or ∂f/∂x are written as ∂xf. Symbols div F, curl F and grad ψ are generally used instead of ∇•F , ∇xF and ∇ψ. The scalar Laplacian is always ∇2. Symbol σ is used for conductivity, so surface charge is relegated to symbol n, which is also used to indicate a derivative normal to a surface ∂nf . Rather than use exotic script fonts or decorations to distinguish various forms of the electric field E, we use an "overloaded" notation where the argument list determines which E function is implied. When an equation is repeated after its first o ccurrence, the equation number is put in italics. A fairly complete list of the symbols used in this document is presented after the summaries below. Probably the length of this list reflects the innate comp lexity of electromagnetic theory and its application to transmission lines. Chapter Summaries Chapter 1 states Maxwell 's Equations and associated equa tions which extend Maxwell's theory from the vacuum to conducting dielectric and magnetic media. After some comments, many of these equations are restated in integral form using the divergence theo rem and Stokes's theorem, and then the behavior of field components at boundaries is obtained. Wave equations for both the fields and potentials are described, and the subject of gauges is dealt with. Starting with Section 1.5 the wave equations are transformed to the frequency domain and Helmholtz equations with parameter β 2 appear. King's Helmholtz integral solutions of these equations ar e then derived using what we call the King gauge. Finally, Section 1.6 clarifies the reasons for using complex fields when physical E and B fields are real. Chapter 2 derives the E and B fields (and current J = σE) inside a round wire which is assumed to have an axially symmetric current flow. The resulting fields are rather complicated and reveal the skin effect. The surface impedance is defined and various quantities are plotted. Assumptions are made about the vector directional nature of the E and B fields in this analysis. The same problem is treated without these assumptions and for an arbitrary transverse current dist ribution in Appendix D. The main results of that lengthy Appendix appear in box (D.4.9). Chapter 3 discusses odd topics such as dielectric loss tange nt, the thickness of surface charge, and why there is no free charge inside a conductor or a dielectric. The chapter concludes with a qualitative Overview and Summary 9 description of the E and B fields of a (TEM) transmission line, with some sketches of the fields. Various Facts about a transmission line are stated. Chapter 4 uses the Helmholtz integral form of the poten tials to derive the well-known transmission line equations which are these, ∂ zV(z) = - z i(z) ∂zi(z) = - yV ( z ) z = R + jωL y = G +jω C . (4.11.14b) In this process, the "transmission line limit" is assume d. It says that the wavelength on the line is much longer than the transverse dimensions of the line. The analysis then yields precise meanings for the parameters R, L, G and C. L is in fact the sum of external and internal inductance contributions L e + Li and it turns out that L e, C and G are all related to each other in terms of a certain dimensionless real parameter K as shown in (4.11.34). Parameters R and L i are the real and imaginary parts of the sum of the conductor surface impedances Z s1 + Zs2. For closely spaced conductors , it is shown how quantities like Zs1 are interpreted as perimeter averages. This chapter's analysis is first carried out assuming that the conductors and dielectric all have the same magnetic permeability μ, but then Section 4.12 shows how to generalize the results for arbitrary magnetic conductors and dielectric. At this point, the transmission line parameters are cl arified and are related to each other, but they are not "known" due to the fact that their solutions involve integral equations over the transmission line geometry. This is the typical chicken-and-egg problem one encounters in all real-world electromagnetic problems. Apart from simple cases (such as ve ry thin transmission line conductors), further approximation must be made. Chapter 5 describes the required approximation. It is basically a continuation of the "transmission line limit" mentioned earlier, along with a notion of "low -loss", which then allows the transmission line problem to be reformulated as a 2D potential theory pr oblem which we call "the transverse problem". It is then basically a "capacitor problem" and then any tr ansmission line geometry can be solved at least numerically. Basically the assumption that the conductors are very good conductors transforms the transverse Helmholtz equation into the 2D Laplace equa tion which is the basis of 2D potential theory. Chapter 6 then gives a complete discussion of the exact so lution, within the assumptions just mentioned, for transmission lines consisting of two parallel solid cylindrical conductors of arbitrary diameter and arbitrary relative (but not intersecting) position. This includes twin-lead lines with equal and unequal conductor diameters as well as on- and off-centered coax ial lines. Since an infinite radius cylinder is a plane, this discussion also obtains the exact soluti on for a transmission line consisting of a round wire over a ground plane. Then the proximity and skin eff ects are analytically calculated for this transmission line and the current density J z is plotted over the wire cross section for various wire sizes, locations, and frequencies. It is shown that J z tracks the charge density n(θ ) around the wire perimeter of each wire. At low ω the entire model is uncertain, and it is shown why the limit ω→0 cannot be interpreted in the way one might think. The active perimeter p and average surface impedance Z s and are then computed. The final section comments on the proximity effect for conductors in which currents flow in the same direction. Overview and Summary 10 Appendix Summaries Appendix A discusses gauge invariance and proves the existence of gauges in which div A can be set to any arbitrary (but reasonable) scalar function, A being the vector potential appearing in B = curl A. The notion of a Green's function or "propagator" is in troduced. A few passing comments are added regarding the connection to special relativity, covariance and quantum field theory. Appendix B analyzes the situation in which the transmissi on line dielectric and conductors have different magnetic permeability μ, a situation not treated in King's TLT book. This causes a bound magnetization current density J m to appear both at boundaries (as a surface current) and in the bulk conductors (as a volume current). It is shown ("the J m theorem") that the theory of Chap ter 4 with its Helmholtz integrals for the potentials can be "rescued" by adding just the surface component of the magnetization current Jm to the true conduction current in the vector potential integrand. A simple method is given for computing this surface Jm current from the conduction current distribution J in the conductor. As usual, the round wire serves as a calculational example. Appendix C concerns the seemingly mundane subject: DC properties of wires . The main issue here is the DC inductance of wires which we treat from a stored energy viewpoint. Internal inductances are computed for a round wire and a hollow round pipe. It is shown that even for a simple rectangular cross section (including square), the internal inductance ca nnot be expressed analytically (at least using our method) and a numerical calculation is required. Th at calculation has been done recently (2009) by Holloway and Kuester. Appendix D computes the E and B fields inside a round wire which is assumed to be one conductor of a transmission line down which a trav eling wave propagates at frequency ω. The solution is obtained using azimuthal partial wave analysis. The E field Helmholtz equations are di rectly solved in cylindrical coordinates, and the B field is then computed from Maxwell's curl E equation. Boundary conditions at the wire surface are discussed. The results (D.4.9) are expressed in terms of the surface charge moment η m in each partial wave. The results for the m=0 partial wave are compared with the results of Chapter 2 which assumed a symmetric current distribution. A passing glance is taken at the corresponding fields outside the wire, then the two boundary conditions are examined more closely, including consideration of Debye surface currents. The source of the surface charge n( θ) is pondered. The dielectric is initially assumed to be non-conducting, but then this restriction is removed. Finally the high ω and low ω limits of the E and B fields are calculated. It is noted that the entire model is not meaningful very close to ω = 0. Appendix E ponders the thickness of the surface charge on a conductor. It is shown that the charge layer thickness is about 1/3 the radius of a copper atom for a copper conductor, and that this is 4000 times smaller than the skin depth at 100 GHz. Appendix F is an elementary discussion of the waveguide modes of a parallel plate transmission line. It shows why there is a cutoff frequency below whic h no waveguide modes can operate, whereas the "transmission line (TEM) mode" on the same structure operates all the way down to very low ω. Appendix G computes the DC vector potential A z inside and outside a round wire carrying uniform current. The computation is done three ways, the most difficult using the Helmholtz (Poisson) integral. When the dielectric surrounding the wire has a μ different from that of the wire, a homogeneous solution Overview and Summary 11 must be added to the Helmholtz particular integral solution. It is this homogeneous solution that is synthesized by adding the surface magnetiza tion current discussed in Appendix B. Appendices H and I derive the Poisson and Helmholtz Green's Functions for the 3D and 2D Poisson and Helmholtz differential equations. These play a major role in the entire document. Appendix J shows how the transmission line transverse analysis replaces 3D propagators with 2D propagators of the Helmholtz and Poisson equations. Results obtained b lindly in the main document are interpreted in terms of these Green's function propagators. Appendix K presents the standard network model of a transmission line as the limit of a set of lumped circuit components. By computing the characteristic impedance Z 0 both from this network model and from Maxwell's equations, it is shown that the R,L,G,C parameters of both models have the same meaning, and this makes the connection between these network-model parameters and those obtained in Chapter 4 from the Maxwell equations. Appendix L considers a point charge located at the center of the cavity of a thick spherical dielectric shell. The problem is solved and limiting cases are obtained. The solution provides an interpretation of how bound charge is accounted for by the dielectric constant ε in E r = (1/4πε) (q/r). This 3D analysis is then repeated in 2D for a line charge in the cavity of an infinite cylindrical shell. Appendix M shows qualitatively that, in the King gauge, the vector potential transverse components A t are much smaller than the longitudinal component A z for all frequencies of transmission line interest. Appendix N describes some subtle aspects of current flow in the presence of magnetic fields. The regular Hall effect is treated, the notion of magnetic Ohm's law is derived, and the Hall effect is reconsidered in light of this law. After dealing with multiple carrier types and magnetoresistance, we show that in a static round wire carrying a current I, the longitudinal current density J z is uniform, and there exists a radial Hall effect inside the wire. There is a small radial electric field E r and a small free charge density ρ inside the wire which is balanced by a small surfa ce charge on the wire surface. The cyclotron frequency ωc plays a major role in this discussion. Appendix O reviews three methods for generating 2D field line plots . The first method is brute force tracking iteration, while the second method makes use of Maple's ability to numerically solve a pair of coupled differential equations. The third analytic me thod works in some cases. An example of each method is presented. Appendix P discusses the eddy current interpretation of the skin and proximity effects. A perturbation expansion is devel oped and for small ω the first term of this expansion is used to compute the eddy currents in some simple cases. A thin round plate is treated analytically for a uniform then for a non- uniform external B field. Then a series of qualitative examples leads to an explanation of the skin and proximity effects in a transmission line as well as in generic parallel wires with same or oppositely directed currents. It is shown why there is current crowding, and why such wires attract or repel. Overview and Summary 12 Appendix Q computes the real and imaginary parts of jk = (R+jωL)(G+jωC) and then evaluates the limits for large and small ω. This task is then repeated for Z 0 = R + jωL G + jωC . Appendix R applies the theory developed in this document to a case study: Belden 8281 coaxial cable. Symbols used in this document Sy mbols are listed "alphabetically" in four groups. Some symbols have multiple meanings separated by semicolons. The list shows the first use location of unusual symbols. The reader seeking entertainment might compare these choices to his or her favorites. Operators and Special Symbols ≡ is defined as ( ≈ approx. equal, ~ ballpark equal) A•B 3D dot product aμbμ implied summation, see for example (1.3.11) a * b regular multiplication a* complex conjugation aka also known as a/b space (and effort) saving version of a b . Examples: ∂F/∂t = ∂F ∂t , sin(x)/(3abσ ) = sin(x) 3abσ (except where expressions are very confusing with the slash notation). ∂t partial time derivative ∂/∂t , so then ∂F/∂t = ∂tF ∂x partial spatial derivative ∂/∂x (similarly ∂y, ∂z, ∂θ, ∂r, etc. ) ∂i partial spatial derivative ∂/∂xi ∂'i partial spatial derivative ∂/∂x'i MT transpose of matrix M QED thus it is proved RHS right hand side, LHS is left hand side Σi=13 budget summation notation (fits on a single line) v• same as d v/dt … d'Alembertian = the 4D version of - ∇2. … = (1/c)2∂t2 - ∇2, see (1.3.11) @ Moon and Spencer notation for the vector Laplacian (also written ∇2) E^ Fourier Transform of E; this ^ notation is used only in Section 1.6 x^ unit vector indicator ∫{ closed line integral, usually ∫{ds Capital Latin A vector potential (1.3.1); A zt is transverse vector potential, see (5.2.1) A a surface area, dA = a differential piece of this area (often dS); A = Angstrom = 10-10 m B magnetic field, see (1.1.5) (sometimes called magnetic induction) B a bipolar coordinate used in Ch 6 (often called ξ elsewhere) C generic constant name (also A, B,C,D...) ; conductor name such as C 1 and C2 Overview and Summary 13 C capacitance (often per unit length of a transmission line) C' complex capacitance, see (4.11.23) C'/C = q c/qs = (ξ/ε) D diffusion constant, see App E and (3.1.1); ge neric transverse dimension of a transmission line D electric displacement, see comment below (1.1.18) E electric field, but see Section 1.6 (f) about our heavily ove rloaded notations G conductance per unit length between conductors of a transmission line H magnetic field, see (1.1.5); H(1)(z) = Hankel function I I = i(0), first used in (D.2.31) (total current in a wire at z = 0) J, J i current density and component thereof Jc conduction current (as opposed to displacement current or polarization current) Jm magnetization current, see (1.1.20) : Jpol is polarization current Jm(z) Bessel function Jμ contravariant 4-vector ( related objects Fμν, ∂μ, ∂μ, Aμ, see App A.6 and Sec 1.3 (b) ) K surface current; constant appearing in transmission line parameter calculations, see (4.4.8) K z surface current component in the z direction KL see (4.10.9) and (4.11.30) Km coefficient appearing in Appendix D L inductance; sometimes a di fferential operator (such as L x = ∂x2 or Lr ) Le external inductance of a transmission line ( does not include energy storage inside conductors) Li internal inductance of a conductor (does not include energy storage outside the conductor) M magnetization, see (1.1.21) Mν(z) Jν(ej3π/4z) = M ν(z) ejθν(z) for real z Nm moment of charge density n( θ) on a round wire, see (D.1.5) P polarization, see (1.1.12) P power, P = IV, see (C.3.3); total conductor perimeter Q total charge on something R resistance; distance between two 3D points (R = | x-x'| ) Rdc DC resistance of a conductor. For a round wire of radius a, R dc = 1/(σπ a2). RH Hall coefficient, see (N.2.3) S a surface area, dS = a differential piece of this area, dS = dS n^ Tz stream function (also T); T is the current vector potential, see (P.3.7) T thickness; temperature; T F = Fermi temperature U energy stored in an inductor, U = (1/2)L I2 , see (C.3.4) V voltage; sometimes volume, diffe rential dV. When mixed, volume is V or dV VH Hall voltage, see (N.2.5) V(z) Δφ between transmission line conductors at some z, see (4.4.1) W(z) ΔAz between conductors at some z, see (4.10.1) W width of something X reactance X C = 1/(ωC), XL ≡ ωLe, see (4.11.34) Zs surface impedance, see Section 2.4 Z0 characteristic impedance of a transmission line, see (4.4.12) and (K.6) Zm intrinsic impedance of a medium = μ/ε , see (4.4.14) Zfs impedance of free space (377 Ω), see (1.1.29) Overview and Summary 14 Zm complex intrinsic impedance of a medium = μ/ξ , see end of Section 4.4 (not used) Lower Case Latin: a radius of a round wire (a 1 and a2 if there are two round wires) am coefficient appearing in Appendix D b distance between centers of a transmission line made of two round wires b(x,y) transverse current density in a conductor, normalized to 1 ber n(z) Kelvin function [also bei n(z)] , Jν(ej3π/4z) = ber ν(z) + j bei ν(z) for real z, see (2.3.1) c speed of light in vacuum ch cosh cof cofactor matrix ( as in c-1 = cof(cT)/det(c) for matrix c ) curl curl (sometimes written as ∇ x ) d diameter of a round wire (d 1 and d2 if there are two round wires); bipolar focal distance dS differential vector S urface area (scalar is dS, but sometimes written as dA) ( A = vector potential) ds differential distance along a curve (written elsewhere as d l ) ; scalar distance is ds dSξ local Stakgold surface area element in n dimensions with ξ the normal direction dV differential volume ; (V = voltage) det determinant of a matrix div divergence (sometimes written as ∇ • ) e electron charge, e = - |e| ; e = 2.71 exp(z) ez f frequency; generic function name f m a combination of Bessel functions, see (D.2.33) gm a combination of Bessel functions, see (D.2.33) grad gradient (sometimes written as ∇ ) g(x|x') Green's function (aka a Green function or propagator); sometimes written g( x,x') or g( x,t; x',t') hm a combination of Bessel functions, see (D.2.33) h height of something, like wire cente r line above a plane; Planck's constant i(z) total current in a conductor at location z, first u sed in (4.7.3) and (4.7.5); i(0) = I in App D. j -1 , see comment above (D.1.3) k sometimes used for a wavenumber (k φ, kA, k in App D, etc. ); Boltzmann's constant k(x,x') kernel in an integral equation or integral expression m meter; partial wave label in Appe ndix D; mass of particle (an electron) n surface charge density; normal vector ( n); normal component (E n); electron density (n e) n(θ) surface charge density on a round conductor which is part of a transmission line n surface charge per unit perimeter distance of a conductor (Cou/m); sometimes per angle nfree free surface charge ns same as n free nc transport surface charge density, see (1.5.17): n c = (ξ/ε)ns p active perimeter length; momentum q generic point charge q(z) total charge per unit length on a conductor at location z, first used in (4.1.2) and (4.1.4) r radial variable for cylindrical coordinates ( ρ is charge density) ; sometimes spherical r s(x) generic source function, see for example (H.1.8) Overview and Summary 15 sh sinh sij a 2D distance between points i and j t time; thickness of something t as subscript means either tangential (E t) or transverse (E t) tanL loss tangent see (3.3.2) ( appears as tan δ in other sources, aka dissipation factor) th tanh u energy density in a magnetic field, u = (1/2) B•H, see (C.3.1) vd speed of light in a dielectric (sometimes just v ) v drift velocity of electrons in a conductor, see App N; sometimes v = generic velocity w width of something x,xa x = β'r, xa = β'a, see (D.2.33) x,x' generic points in 3D space (sometimes written r,r') x^ unit vector in the x direction (similarly y^, θ^, n^ etc.) y admittance per unit length of a transmission line see (4.11.14b) z impedance per unit length of a transmission line see (4.11.14b) z longitudinal dimension of a transmission line or round wire; generic Bessel Function argument Greek // pseudo-alphabetical α(x,y) transverse charge density in a conductor, normalized to 1 (is delta function on surface) β wavenumber for transmissi on line conductor, see (1.5.1c) βd wavenumber for dielectric surrounding transmission line conductors, see (1.5.1a) βd0 wavenumber for non-conducting dielectric, see (1.5.1b) β' β'2 ≡ β2 - k2 see (D.2.2) ∂ partial derivative (see operator list above) δ(x) Dirac delta function; δ(r) = 3D delta function δ skin depth, see (2.2.20) δi,j Kronecker delta ε absolute electric permeability ( ε = ε0 in vacuum), ε = ε' - jε"; a small real quantity ε>0 εrel ε/ε0 εijk permutation tensor ξ complex electric permeability (1.5.1) ( ξ = ε - σ/jω); ξ = a bipolar coordinate called B in Ch. 6 ξ Stakgold n-1 dimensional coor dinate of a point on a surface σ φ scalar potential, see (1.3.1); φt is the transverse scalar potential, see (5.1.1) κ = 1/μ , inverse mobility, see (N.8.1) λ wavelength λD Debye length, see Appendix E Λ generic gauge function (Appendix A.2); arbitrary large cutoff value, see (J.10) ηm normalized moment of charge density on a round wire, see (D.2.30) π 3.14 ( = Pi in Maple V) ρ charge density (Cou/m3); resistivity ρ = 1/σ (ohm-m); ρfree free charge density ρpol polarization charge density, see (1.1.11) σ conductivity (surface charge therefore is n, not σ) ; standard deviation; Stakgold surface label σeff effective conductivity see (3.3.4) Overview and Summary 16 θ azimuthal angle for cylindrical coordinates ( φ is scalar potential) θν(z) J ν(ej3π/4z) = M ν(z) ejθν(z) for real z θ(a>b) Heaviside step function, normally written θ(a-b) or H(a-b) τ collision time, see Appendix N μ absolute magnetic permeability ( μ=μ0 for vacuum) ; carrier mobility in Appendix N; mean value ω angular frequency ωc cyclotron frequency, see (N.3.1); soft low cutoff frequency, see (D.11.11) χe,χm electric and magnetic susceptibility Chapter 1: Basic Equations 17 Chapter 1: Basic Equations In this chapt er we state the basic equations to be used later in the calculation of transmission line parameters and in the exploration of transmission line behavior. 1.1 Maxwell's Equations in a Conducting Dielectric Medium Our working set of equations is the following: curl H = ∂tD + J Maxwell curl H equation ( J = Jc) (1.1.1) curl E = - ∂ tB Maxwell curl E equation (1.1.2) div D = ρ Maxwell div D equation ( ρ = ρfree) (1.1.3) div B = 0 Maxwell div B equation (1.1.4) B = μH magnetic permeability μ (1.1.5) D = εE electric permeability ε (dielectric constant) (1.1.6) J = σE O h m ' s L a w ( σ = conductivity) (1.1.7) div J = - ∂ tρ Equation of Continuity (see item 7 below) (1.1.8) (a) Notes on Maxwell's Equations Although Maxwell 's equations ("th e Maxwell equations") provide a concise overview of classical electrodynamics, there is lot going on "under the hood" and clarification of the meaning of certain symbols seems useful, hence the following set of notes. 0. It is understood that, in a medium other than th e vacuum (that is, a "ponderable" medium), all the mathematical fields shown above like E, D, B, H, J, ρ (and later A and φ) are average fields in the sense discussed in Jackson Sections 4.3 and 6.6. The pa rtial differential equations are meaningful for differential volumes, areas and distances which are very small but still contain enough atoms or molecules (perhaps at least 1000) so that averaging makes sense. We shall refer to the various electric and magnetic fields as "fields" to distinguish them from "potentials" like A and φ, though all these quantities are mathematical fields. 1. The equations above are all expressed in SI units. The connection with cgs/Gaussian units is explained in an Appendix present in all three of the Jackson Classical Electrodynamics editions. The above equations appear in Jackson's third edition at these locations, Chapter 1: Basic Equations 18 (1.1.1) through (1.14): p 2 (I.1a) Maxwell's Equations (1.1.5) and (1.16): p 296 top line permeability constitutive relations (1.1.7) p 219 (5.159) Ohm's Law constitutive relation J = σE (1.1.8) p 3 (I.2) equation of continuity 2. All media (conductors, dielectrics between conducto rs) are assumed to be homogeneous and isotropic so that the quantities σ, ε and μ are constant scalars in space (not tensors) for a given medium. In the vacuum these constants take the values σ = 0, ε = ε 0 and μ = μ0. An implication of σ, μ and ε being constants in space is that they pass through the div, curl, grad and ∇2 operators just as would any constant like π. One must be a little careful at a boundary between homogenous media since these constants can be different in the two media. In principle, all three qua ntities can vary in time, an d when transformed to the frequency domain, σ(ω), ε(ω) and μ(ω) can (and do) vary with ω. However, we shall assume that for our frequencies of interest, these quantities are constant in ω and are therefore also constant in time so they pass through ∂t. 3. The difference between ε and ε0 is caused by polarization of bound charge in a medium. Equations dealing with polarization are these [ see Jackson pp 1 53-4 or Panofsky & Phillips pp 28-30 and p 129-130 on the polarization current ] : PdV = electric dipole moment contained in volume dV of a dielectric (1.1.9) J pol ≡ ∂tP = polarization current density (1.1.10) ρ pol = - div P = polarization charge density (1.1.11) P = ε 0χeE // polarization assumed proportional to the polarizing E field (1.1.12) D = ε 0E + P = ε0(1 + χe)E = ε E = "the electric displacement " (1.1.13) ε = ε 0(1 + χe) // ε = dielectric constant, χe = electric susceptibility (1.1.14) div E = (1/ε 0)(div D - div P) = (1/ε 0)(ρfree + ρol) . " E sees all charges" (1.1.15) The E field causes polarization P either by causing existing tiny dipole objects (e.g., molecules) in a medium to "line up", or by causing tiny non-dipole objects (e.g., atoms) to have dipole moments and then those get lined up. See for example Bleaney & Bleaney Chapter 10 " Dielectrics". Comment : Since D = ε0E + P, the D and E fields are scaled differently. It might have been better had the D field been replaced by D = ε0D' in which case D' = E + P/ε0; then one can make clearer statements about D' versus E. For example, in a dielectric capacitor with fixed conductor charges (Q,-Q) there exist both D' and E fields, and D' = (ε /ε0) E > E. The D' field can be interpreted as the E field that would be present were the dielectric replaced by empty space. The dielectric in effect shields the charge, reducing E and hence V, does not change Q, and, since Q = CV, it increases capacitance C by ( ε/ε0) for fixed Q. Chapter 1: Basic Equations 19 4. As noted in (1.1.3), the ρ in div D = ρ is the free charge density ρ free and does not include possible polarization charge density. In contrast, the E field "sees" both free charge ρfree and polarization charge ρpol , as derived above in (1.1.15) from (1.1.13), div E = (1/ε0) ( ρfree + ρpol) = (1/ε) ρfree . (1.1.16) In the rightmost expression, the polarizati on charge is incorporated into the 1/ ε factor. Appendix L shows how this works physically in the case of point and line charges embedded in a dielectric. 5. The J in curl H = ∂ tD + J is the conduction current Jc . If polarization current Jpol is present, it is included in the "displacement current" term ∂tD along with the Maxwell "vacuum polarization current" Jvac = ε0∂tE . That is, J d ≡ ∂tD = ∂t[ε0E + P] = ∂tP + ε0∂tE = Jpol + Jvac ρpol = - div P . (1.1.17) The Jvac term ε0∂tE was "added" by Maxwell to the curl H equation (Ampere's Law) to make it self- consistent. Since div curl H = 0, and since curl H = Jd + Jc, one must have div [ Jd + Jc] = 0 : div [ Jd + Jc] = div [∂tP + ε0∂tE] + div[ Jc] = ∂t[div P + ε0 div E] - ∂tρfree = ∂ t(-ρpol) + ∂t(ρfree + ρpol) - ∂tρfree = 0 ( 1 . 1 . 1 8 ) where we have used continuity div J c = -∂tρfree, see item 7 below. Comment on "Displacement" : In the case of polar molecule polarization, the polarization charge and current can be viewed as being caused by a "displace ment of bound charge" as suggested by this very symbolic picture of a parallel plate capacitor Fig 1.1 The applied E field of the plates lines up the polar molecules and thus causes a polarization charge density n pol to appear on the side faces of the dielectric, as if it were an "electret" object. One can imagine that, with an AC plate voltage, as the applied E field changes to the other polarity, the polar molecules rotate in place 180 degrees putting the pos itive bound charge on the opposite plate, and as this happens, there is a polarization current Jpol = ∂tP inside the dielectric. In reality, the molecules are close Chapter 1: Basic Equations 20 to randomly oriented and the above effect is obtained for the "average" molecule. In any event, the E field causes surface polarization charge densities n pol at the faces of the dielectric, and one then thinks of the normally neutral-everywhere bound charge distribu tion as being "displaced" such that one face has positive charge and the other negative. It is in this sense that Maxwell started using the word "displacement". Before the Jvac term was added, Maxwell had Jd ≡ ∂tD = Jpol and this associated ∂tD entirely with Jpol and thus with the displacement of the di electric bound charge, and so Maxwell referred to D as the "electric displacement" and ∂tD as the "displacement current". Fig 1.1 shows how the polarization charge acts to shield the free charge, so n tot = nfree - npol. 6. If there is any magnetization current J m = curl M , it is absorbed into the distinction between B and H and therefore does not appear on the right side of curl H = ∂tD + Jc . The current Jm is discussed for example in Panofsky & Phillips, Sections 7-12, 7-13 and 8-1. The basic equations are as follows, MdV = magnetic dipole moment contained in volume dV of a medium (1.1.19) Jm = curl M = magnetization current density ( => div Jm = 0) (1.1.20) M = χ m H = magnetization // = [ μ/μ0- 1] H from (1.1.23) (1.1.21) B = μ 0(H+M) = μ0(1+χm)H = μH = "magnetic induction" (infor mally, magnetic field) (1.1.22) μ = μ0(1+χm) // μ = magnetic permeability, χm = magnetic susceptibility (1.1.23) curl B = μ0(curl H + curlM ) = μ0(∂tD + Jc) + μ0Jm = μ 0(∂tD + Jc + Jm) . " B sees all currents" (1.1.24) 7. The "equation of continuity" (1.1.8) expresses the fact that charge cannot be created or destroyed. Barring ionization of a dielectric, free charge and bound charge (polarization charge) cannot be converted into each other and are therefore separately conserve d. Thus we have several different equations of continuity: [ see for example Haus and Melcher, Section 6.2, equations (10) and (13) ] div J c = -∂tρfree // conservation of free charge (aka true or unpaired charge) (1.1.25) div Jp = -∂tρpol // conservation of polarization charge (a ka bound or paired charge) (1.1.26) div [ Jc+ Jp] = -∂t[ρfree + ρpol] = -∂tρtot // sum of above two equations (1.1.27) div [ Jc+ Jd] = 0 ≠ -∂tρtot // reminder of item 5 above (1.1.18) 8. Ohm's Law J = σE is assumed to be a valid constitutive relation for our media of interest. One should keep in mind that this is an approximation, wher eas the Maxwell equations and the continuity equations are not. Just under the surface charge on a conductor, Ohm's Law is violated as discussed in Appendix E due to a diffusion current generated by charges piled up at the surface. Oh m's Law is also violated in the Chapter 1: Basic Equations 21 presence of very strong magnetic fields as shown in (N.4.10). In this case one can say that Ohm's Law is still valid, but σ is a tensor instead of a scalar. 9. As will be shown in Section 3.1, inside a medium such as a dielectric or a conductor, and at frequencies of interest to us, there can exist no net charge densities, so ρ free = 0. In a dielectric there are no available free charges, while in a conductor, any departure from neutrality would be instantly restored. All free charge densities for our application reside on the surfaces of conductors only. If we were interested in the behavior of a transmission line embedded in an charged plasma, things would be different. 10. As noted in item 1, all our equations are expressed in Système Internationale (SI) units. In this system, formerly known as "rationalized m.k.s.", the speed of light is concealed in the symbols μ 0 and ε0. Here are the usual historical names given to the symbols appearing in our equations, along with one expression of the SI units for each symbol: E = electric field (volts/m) H = magnetic field (amp/m) D = electric displacement (coulomb/m 2, same units as surface charge) B = magnetic field (tesla = amp-henry/m2 = volt-sec/m2 = weber/m2) 1 tesla = 10,000 gauss J = current density (amps/m2) ρ = charge density (coulombs/m3) σ = conductivity of the medium (mho/m = ohm-1/m) μ/μ0 = relative magnetic permeability of the medium (dimensionless) ε/ε0 = relative electric permittivity = relativ e dielectric constant (dimensionless) μ0 = permeability of free space = 4 π x 10-7 henry/m ε0 = permittivity of free space = 8.8541877 x 10-12 farad/m (1.1.28) Here are some unit relations obtainable from Q = CV, V = IR, LC = 1/ ω2 , τ = RC = L/R, I = dQ/dt : coulomb = farad-volt volt = ampere-ohm henry-farad = sec 2 farad = sec/ohm henry = ohm-sec henry / farad = ohm2 ampere = coulomb/sec mho = ohm-1 mho/F = sec-1 newton = coulomb-volt/m = kg-m/sec2 // F = qE = ma amp-henry = volt-sec c = 1/ μ0 ε0 = 2.9979246 x 108 m/sec = speed of light Zfs = μ0/ε0 = 376.73032 ohms = "impedance of free space" σ = 5.81 x 107 mho/m for copper (1.1.29) Notice how the names of eight people have become forever embedded into the SI unit system. Comment: Inevitably, any given author will at some point refer to both B and H as "the magnetic field". We shall do that throughout, using the historical symbols B or H to indicate which "kind" of magnetic Chapter 1: Basic Equations 22 field we are talking about. Some authors refer to B as the magnetic flux density or the magnetic induction to distinguish B from H. (b) Integral Forms of Maxwell's Equations and Continuity The equations above invol ving the divergence and curl operators have integral forms thanks to these two fundamental mathematical theorems which have nothing to do with electromagnetism in particular, ∫V div F dV = ∫S F • dS // "the divergence theorem" Spiegel 22.59 (1.1.30) ∫S curl F • dS = ∫{C F • ds // "Stokes's theorem" Spiegel 22.60 (1.1.31) F i g 1 . 2 The first theorem involves a closed boundary surface S which encloses a volume V and says that the volume integral of div F over V equals the surface integral of F over S. The second involves a closed bounding curve C (possibly non-planar) which bounds an arbitrary open surface S (also possibly non-planar) and says that the line integral of F around C equals the surface integral of curl F over S. In the divergence theorem, d S points "out" from the volume, and in Stokes's Theorem, the direction of d S and d s are related by the right-hand rule where fingers f it the boundary curve and the thumb gives the direction of dS. In both theorems the differential vector area patch is d S = dS n^ where n^ is normal to the surface. Both theorems have meanings in n-dimensional space , but our interest is mainly n = 3. Both theorems are not hard to derive and this is done in textboo ks usually by breaking up the surface into tiny squares and the volume into tiny cubes. Once one sees these derivations, the theorems become less mysterious. In general terms, the divergence theorem says that div F is somehow a source of the field F and the amount of F flowing out through a closed bounding surface equals the amount of F that is generated inside the volume. When F is the electric field E, the divergence theorem is called Gauss's Law and says Chapter 1: Basic Equations 23 that the total electric flux "flowing out" [ that is to say, ∫S E•dS ] equals the total of the source inside the volume [ (1/ε )∫V ρ dV ], usually called "the to tal charge enclosed". Thus, div D = ρ ⇔ ∫V ρ dV = ∫S D • dS ( 1 . 1 . 3 2 ) div E = ρ/ε ⇔ ∫V ρ dV = ∫S ε E • dS . (1.1.33) Since the magnetic field has no corresponding charge, one always has ∫S B • dS = 0, a theorem which seems to have no name, div B = 0 ⇔ ∫S B • dS = 0 S is any closed surface (1.1.34) The surface integral of an E or B field is often referred to as the total electric or magnetic "flux" passing through the surface, even though nothing is really flowing in a mechanical sense. The divergence operator also occurs in the equation of continuity (1.1.8) so we have div J = - ∂ tρ ⇔ -∂t[∫V ρ dV] = ∫S J • dS . (1.1.35) This is the prototype application of the divergence th eorem in that it is easily understandable: the total electric current flowing out through some closed surface S must equal the rate at which the total charge inside the surface is decreasing. On e can write a similar statement for mass flowing out from a volume in which ρ would be the mass density and J = ρv the mass current, v being the velocity field. The Stokes theorem is a bit more mysterious. Since th is theorem is associated with George Stokes, it is called Stokes's theorem, but is sometimes called Stok es' theorem (one would not say Gauss' theorem). The curl of a vector field is associat ed with the amount of "rotation" the field has at some point in space, and in fact curl is sometimes written Rot. If one cons iders a tiny patch and finds that the line integral of the field around the boundary of that patch is non-zero , then the vector field has a non-zero curl at that point in the direction normal to the patch. At any point where a fluid has a vortex, the curl is non-zero, for example. When Stokes's theorem is applied to the electric field, one has curl E = - ∂ tB ⇔ ∫{C E • ds = -∂t[∫S B • dS] . (1.1.36) This says that the voltage induced around a closed l oop (the "electromotive force") is proportional to the rate of change of the magnetic flux through that loop, a principle known as Faraday's Law of Induction . If water power rotates a wire loop in the presence of some magnets, one has an electric generator. Chapter 1: Basic Equations 24 On the other hand, when Stokes's theorem is applied to the magnetic field, one gets curl H = ∂tD + J ⇔ ∫{C H • ds = ∫S [∂tD+J] • dS (1.1.37) curl B = με ∂tE + μJ ⇔ ∫{C B • ds = μ ∫S [ε ∂t E + J] • dS (1.1.38) μ constant in space, ε constant in time When the situation is static, one has ∫{ H • ds = ∫S dS • J which says the line integral of the magnetic field H around some loop equals the total current pa ssing through any open surface whose boundary is that loop (the "current enclosed"), a principle known as Ampere's Law . Later we shall encounter a certain "vector potential A" which is related to the B field by B = curl A. Since we are writing out "integral forms" of differentia l relationships, we can then add this to the list, curl A = B ⇔ ∫{C A • ds = ∫S B • dS . (1.1.39) If the bounding curve C were a wire carrying a current I which creates both A and B, then both sides of the above integral form will be propor tional to I, and the constant of proportionality is by definition the self-inductance L of the loop, ∫{C A • ds = ∫S B • dS = [magnetic flux through surface S] = L I . (1.1.40) There are of course many surfaces S which span a give n curve C, and (1.1.39) says that all such surfaces give exactly the same ∫S B • dS and thus the same L, so L is really a geometric property of the curve C. We shall be using all these integral forms in the document below. (c) Rules for behavior of fields and potentials at a boundary Consider the boun dary between two different media ca lled 1 and 2. Consider a tiny red "math loop" of width L and height 2s which straddles the media boundary which here is seen edge on, Fig 1.3 Chapter 1: Basic Equations 25 For the electric field we have from above (for our loop, d S = dS z^) curl E = - ∂tB ⇔ ∫{ E • ds = - [∫S (∂tB) • dS] . (1.1.36) Since the loop is tiny and since the fields are assumed to be non-singular, we can regard E and B as a constant everywhere on each half of the loop (for our purposes here). The line integral around the loop is then (start at lower left corner) ∫{ E • ds = LEx(2) + s Ey(2) + s Ey(1) – LEx(1) - s Ey(1) - s Ey(2) = L [E x(2)- Ex(1)] . The area integral on the right side of (1.1.36) is -∫S (∂tB) • dS = - ∂tBz(1) sL - ∂tBz(2) sL = - sL [ ∂tBz(1) + ∂tBz(2)] so the integral form in (1.1.36) says [E x(2)- Ex(1)] L = - s L [∂ t Bz(1) + ∂t Bz(2)] . As long as ∂ tBz is finite at the surface, as s →0 the right side vanishes and we conclude that [E x(2)- Ex(1)] = 0 . We then summarize for both the x and z directions by saying (t means tangential to boundary) E t1 = Et2 or (1/ ε1)Dt1 = (1/ε2)Dt2 (1.1.41) so the tangential (parallel) components of the electric field is continuous through a boundary. A similar analysis using the curl H equation, curl H = ∂ tD + J ⇔ ∫{ H • ds = ∫S [∂t D+J] • dS (1.1.37) leads to [H x(2)- Hx(1)]L = s L[ ∂t Dz(1) + Jz(1) + ∂t Dz(2) + Jz(2)] . As long as ∂ tDx and Jx are finite (non-singular) at the surface, we conclude from s →0 that Ht1 = Ht2 o r ( 1 / μ1)Bt1 = (1/μ2)Bt2 . (1.1.42) Chapter 1: Basic Equations 26 However, it is possible to have J be singular at the surface in the form of a surface current K where J = K δ( y ) J = a m p / m2 K = amp/m (1.1.43) and in this case we find that [Hx(2)- Hx(1)] L = ∫S [J] • dS = ∫S Kδ(y) • dS = ∫S Kz δ(y) dx dy = ∫S Kz dx ≈ Kz L which we summarize as H t2 - Ht1 = Kzfree or (1/ μ2)Bt2 - (1/μ1)Bt1 = Kzfree (1.1.44) where we imagine z^ = t^ x n^ as the meaning of the z in K z. Notice that this K z is a "free" surface current, and not a bound magnetization surface current since such a magnetization current is not "seen" by H. We mention here a result similar to (1.1.44) which a pplies to a special situation of Fig 1.3 above where we assume a vector potential of the form A = Az(x,y) z^ . This vector potential is constant on the boundary surface in the x direction, and has only an A z component. In this case, B = curl A = x^ (∂yAz - ∂zAy) + y^ (∂zAx - ∂xAz) + z^ (∂xAy - ∂yAx) = x^ (∂yAz) = x^ Bx where B x = ∂yAz (1.1.45) and then (1.1.44) says (1/μ2) (∂nAz)2 - (1/μ1) (∂nAz)1 = Kzfree ( 1 . 1 . 4 6 ) where ∂ n is the derivative of the vector potential A z in a direction normal to the surface (n pointing from medium 2 to medium 1). If μ1= μ2= μ0, we have ( ∂nAz)2 - (∂nAz)1 = μ0Kz and then K z is proportional to the normal slope jump in A z at the boundary surface. As earlier, K z is a "free" surface current. Next, we put a tiny "Gaussian pillbox" straddling the two media. Area A and height 2s are both very small . Gaussian Pillbox a pill box circa 1830 Chapter 1: Basic Equations 27 F i g 1 . 4 For the electric displacement D we consider div D = ρfree ⇔ ∫V ρ dV = ∫S D • dS (1.1.13) where volume V is of the box shown. The surface integral is ∫S D • dS = Dy(1)A - Dy(2)A + contributions from the sides of the box Since we assume D is non-singular, the side contributions vanish as s → 0 since the side area vanishes. Assuming a charge density n free exists on the boundary between the two media, the volume integral is nfreeA and then the conclusion, generalized to the perpendicular field component, is Dn1 - Dn2 = nfree or [ε1E1n - ε2E2n] = nfree . (1.1.47) where n^ points into medium 1. If the two media are conducting dielectrics with Ohm's law Jc = σE, we can apply continuity (1.1.25) to the Gaussian box to find that div Jc = - ∂tρfree ⇔ -∂t[∫V ρfree dV] = ∫S Jc • dS (1.1.35) so that -∂tnfree = [Jn1- Jn2] = σ1En1 - σ2En2 or for monochrome time dependence (coming soon, along with notation explanation), -jω nfree = σ1En1 - σ2En2 . // frequency domain Recall now from (1.1.47) that n free = ε1En1- ε2En2 . (1.1.47) Adding the last equation to 1/j ω times the previous equation gives 0 = [ε1 + σ1/jω] En1 - [ε2 + σ2/jω]En2 In terms of the complex dielectric constants ξ i ≡ εi + σi/jω this says that 0 = ξ1En1 - ξ2En2 so that ξ 1En1 = ξ2En2 // frequency domain (1.1.48) In the limit that, say, medium 2 becomes a perfect conductor, ξ 2 ≈ σ2/jω → ∞ and En2 → 0, but the product is maintained equal to ξ1En1 . Chapter 1: Basic Equations 28 Returning again to the special case in which vector potential A = Azz^ , since E = - ∇φ - ∂tA (as shown later), we have E y = -∂yφ since A y = 0. In terms of Fig 1.3 where y is the direction normal to the surface, one has E n = -∂nφ and then (1.1.47) may be written ε2(∂nφ)2 - ε1(∂nφ)1 = nfree ( 1 . 1 . 4 9 ) which can be compared to (1.1.46). If ε 1 = ε2 = ε0, we have ( ∂nφ)2 - (∂nφ)1 = nfree/ε0 and then n free is proportional to the normal slope jump in φ at the boundary surface. Finally, the other divergence equation div B = 0 ⇔ ∫S B • dS = 0 S is any closed surface (1.1.34) leads to the conclusion that B n1 = Bn2 o r μ 1Hn1 = μ2Hn2 . (1.1.49) We now summarize these rules in a box, always assuming that there is no singularity in some quantity to invalidate the claims: Rules for continuity of normal and tangential fields at a boundary: (1.1.50) The fields here are either F(x,t) or F(x,ω), except(1.1.48) which is only for E(x,ω) : t = tangential = parallel = || : Et1 = Et2 or (1/ ε1)Dt1 = (1/ε2)Dt2 (1.1.41) Ht2 - Ht1 = Kzfree or (1/ μ2)Bt2 - (1/μ1)Bt1 = Kzfree (1.1.44) Special case A = Az(x,y) z^ : (1/μ 2) (∂nAz)2 - (1/μ1) (∂nAz)1 = Kzfree (1.1.46) n = normal = perpendicular = ⊥ : ( symbol n is also used for surface charge density) Bn1 = Bn2 o r μ 1Hn1 = μ2 Hn2 (1.1.49) Dn1 - Dn2 = nfree or [ ε1En1 - ε2En2] = nfree (1.1.47) and for monochrome time dependence: ξ1En1 = ξ2En2 where ξ = ε + σ/jω (1.1.48) Special case A = Az z^ : ε2(∂nφ)2 - ε1(∂nφ)1 = nfree (1.1.49) Chapter 1: Basic Equations 29 Tangential and normal boundary conditions can always be written in the following manner, F t1 = Ft2 ⇔ n x F1 = n x F2 Fn1 = Fn2 ⇔ n • F1 = n • F2 ( 1 . 1 . 5 1 ) as can be seen by expanding F = Fnn^ + Ftt^ and noting that n^ x t^ = 0 and n^ • n^ = 1. The E and B boundary conditions in the above table appear as fo llows in King (1945), page 204 (obtained from the University of Utah's robotic automated retrieval center ARC), where n^1 = - n^2, (n,F) = n • F , [n,F] = n x F , and ν = 1/μ. 1.2 The Field Wave Equations In the following, quantities μ and ε are treated as constants, independent of space and time. The E wave equation may be derived using these steps : curl E = - ∂ tB // Maxwell (1.1.2) curl curl E = -∂ tcurl B = - μ∂t[curl H] = -μ∂t[ ∂tD + J ] // curl both sides and Maxwell (1.1.1) grad div E - ∇2E = -μ∂t[ ∂t[εE] + J ] // vector identity on left and D = εE ( ∇ 2 - με∂t2)E = μ∂tJ + (1/ε) grad ρ . // div E = ρ/ε The B wave equation uses these steps : curl H = ∂tD + J // Maxwell (1.1.1) curl curl H = curl [∂ tD] + curl J // curl both sides grad div H - ∇2H = ε ∂t(curl E) + curl J // vector identity on left and D = εE (1/μ)grad div B - ∇ 2H = εμ ∂t(-∂tH) + curl J // Maxwell (1.1.2) and B = μH twice ( ∇ 2 - με ∂t2)H = - curl J // since div B = 0 (1.1.4) Chapter 1: Basic Equations 30 The two results are then ( ∇ 2 - με ∂t2)E = μ∂tJ + (1/ε) grad ρ ( 1 . 2 . 1 ) ( ∇2 - με ∂t2)B = - μ curl J ( 1 . 2 . 2 ) which agree with Jackson p 246 (6 .49) and (6.50). Recall that με = 1/v2 where v is the speed of light in the medium of interest. These two equations are undamped driven wave equations. 1.3 The Potential Wave Equations (a) The Potential Wave Equations in the Lorenz gauge It is possible to work with the scala r and vector potentials φ and A instead of the fields E and B. If φ and A can be determined, then E and B are fully determined by (1.3.1) belo w. However, in the other direction, if E and B are known, then φ and A are determined only up to a certain "gauge transformation" degree of freedom, a subject discussed in Appendix A. The fields E and B are physically observable quantities while the potentials φ and A in general are not and s hould be regarded as intermed iate "helper" functions. In SI units, the E and B fields are obtained from φ and A in this manner : [ Jackson p 239 (6.7) and (6.9)] B = curl A E = - grad φ - ∂tA . (1.3.1) A = vector potential (tesla-m = amp-henry/m = volt-sec/m) E = volt/m φ = scalar potential (volts) B = tesla . Appendix A (Fact 4) shows that there is a continuum of possible choices ( φ,A) all of which give the same physical fields ( E,B) according to (1.3.1). It turns ou t that, along this continuum, div A takes different functional forms. Fact 4 shows that there always exists a choice ( φ,A) for which div A = any function one wants! Selecting f(x) for div A = f(x) is called "making a gauge c hoice". Different gauge choices just result in different ( φ,A) potentials, but always the same ( E,B). In the following derivations of the wave equations for A and φ, we shall be making a certain gauge choice as indicated. The following steps are used to develop the φ wave equation. In the vacuum one has μ = μ 0 and ε = ε0 and με = 1/c2 and these are the parameters one sees in the Jackson equation references below. E = - grad φ - ∂ tA // (1.3.1) [= Jackson (6.9)] div E = - div grad φ - ∂ t (div A) // take div of both sides ∇2φ + ∂t[div A] = -ρ/ε // div E = ρ/ε [= Jackson (6.10)] (1.3.2) ( ∇ 2 - με ∂t2)φ = - (1/ε)ρ . // apply gauge choice divA = - με ∂tφ Chapter 1: Basic Equations 31 And the following steps are used to develop the A wave equation: curl H = ∂tD + J // Maxwell (1.1.1) (1/μ) curl curl A = με ∂ tE + J // H = B/μ , B = curl A from (1.3.1), and D = εE grad div A - ∇2A = με ∂t[- grad φ - ∂tA] + μJ // vector identity and (1.3.1) E = - grad φ - ∂tA (∇2 - με ∂t2) A = grad [με ∂tφ + div A ] - μJ // [ = Jackson (6.11) ] (1.3.3) ( ∇2 - με ∂t2)A = - μJ . // apply same gauge choice divA = - με ∂tφ The results are then ( ∇ 2 - με ∂t2)φ = - (1/ε)ρ [ = Jackson (6.15) ] (1.3.4) ( ∇2 - με ∂t2)A = - μJ [ = Jackson (6.16) ] (1.3.5) div A = - με ∂tφ . [ = Jackson (6.14) ] // Lorenz Gauge (1.3.6) As discussed in the comment below, this gauge c hoice is now called the Lorenz Gauge. One should notice how the gauge choice decouples the two wave equations (1.3.2) and (1.3.3) so one resulting equation only involves φ and ρ, while the other involves only A and J. We end up then with undamped driven wave equations with simple driving terms. Note that ρ = ρfree (does not include polarization charge ρpol) and that J = Jc (does not include magnetization current Jm). In effect, ρ pol and Jm are incorporated into the constants ε and μ. Comment 1 : For perhaps 100 years pretty much all ( non-Danish) papers and textbooks (including Jackson's first two editions in 1962 and 1975 and th e initial six printings of his 1998 third edition) referred to the Lorenz gauge as the Lorentz gauge, and it was then convenient to say that the Lorentz gauge condition is Lorentz invariant since it transforms as a scalar equation under Lorentz transformations. Now we have to say that the Lore nz gauge is Lorentz invariant because Lorentz was mistakenly credited for first using this gauge conditio n, see Jackson's note p 294 added in his 7th printing. Although the Dane Ludvig Lorenz (1829-1891) was 24 y ears older than the Dutchman Hendrick Lorentz (1853 –1928), they were contemporary though independent workers at the time (1867) that Lorenz first published the use of his now-eponymous gauge condition. Lorentz will just have to be content with his transformations, his invariance, his contraction and his force law which says F = q( E + v x B). For more on Lorenz and Lorentz, see Nevels and Shin. Comment 2: We speak of (1.3.6) as "the Lorenz gauge" and div A = 0 as "the Coulomb gauge". These gauges are really conditions on A and do not fully specify A since many vector fields A can have the same divergence. So a gauge specifies a class of possible A fields, not a particular one. Chapter 1: Basic Equations 32 (b) Special Relativity Note At first encounter, one is amazed at how similar th e two equations (1.3.4) and (1.3.5) appear. Here we shall show why that is. We now assume the medium is the vacuum so με = μ0ε0 = 1/c2. Then the two equations may be written ( ∇2 - 1 c2 ∂t2)φ = - (1/ε 0)ρ ( 1 . 3 . 7 ) ( ∇2 - 1 c2 ∂t2)A = - μ0J . ( 1 . 3 . 8 ) As shown in Appendix A.6, one can construct Lorentz 4-vectors Aμ = ( 1 c φ, A) and Jμ = (cρ , J) with the identification of A0 ≡ φ/c and J0 ≡ cρ. The above equations can then be written, using proper tensor notation where common vectors are contravariant with an upper index, ( ∇2 - 1 c2 ∂t2)[cA0] = - (1/ε 0)[J0/c] = - (1/ ε0)[J0/c] (c2μ0ε0) = - μ0 cJ0 (1.3.9) ( ∇2 - 1 c2 ∂t2)Ai = - μ0Ji . ( 1 . 3 . 1 0 ) Cancelling the c's in the first equation allows both equations to be written as a single 4-vector equation (∇2 - 1 c2 ∂t2)Aμ = - μ0 Jμ or … A μ = μ0 Jμ where … ≡ ∂μ∂μ = 1 c2 ∂t2 - ∇2 . (1.3.11) This equation is covariant because both sides tr ansform as a Lorentz 4-vector (the operator … transforms as a Lorentz scalar). Special relativity requires that all equations of physics be covariant under Lorentz transformations. This is similar to Newton's Law F = m a being covariant under rotations, where both sides transform as 3-vectors. If we start with the correct law of physics (1.3.11) and work backwards through the equation pairs above, where we add a medium with μ and ε, we end up with our starting point (1.3.4) and (1.3.5 ) and the similarity of these two equations is then explained as being a requirement of special relativity. Recall the Lorenz gauge choice (1.3.6) whic h was required to decouple things above, div A = - μ 0ε0 ∂tφ = - 1 c2 ∂tφ . (1.3.6) As shown in Appendix A.6, this Lorenz gauge condition can be expressed in covariant form as ∂μAμ = 0 ⇔ div A = - 1 c2 ∂tφ ( 1 . 3 . 1 2 ) Chapter 1: Basic Equations 33 while the equation of continuity states ∂μJμ = 0 ⇔ div J = -∂tρ . (1.3.13) Both sides of these last two tensor-notation equa tions transform as a rank-0 tensor (scalar) so the equations are covariant (0 is a scalar). As one cha nges frames of reference doing Lorentz transformations (rotations and "boosts"), the potential wave equation, the gauge condition, and the continuity relation always maintain the same tensor form. In closing this relativity note, we must me ntion that the four Maxwell equations (with ε = ε0 and μ = μ0 and μ0ε0 = 1/c2) can also be stated in covariant notation. One first defines the following antisymmetric rank-2 tensor (see Appendix A.5 concerning up and down indices etc.) Fμν ≡ ∂μAν - ∂νAμ Aμ = ( 1 c φ, A) Jμ = (cρ, J) ∂μ = (∂0, ∂i) = (∂0, -∂i) (1.3.14) where obviously Fμν = -Fνμ and Fμμ = 0 for diagonal elements. Then the two Maxwell homogeneous (no sources) equations appear as ∂αFμν + ∂νFαμ + ∂μFνα = 0 // both sides transform as a rank-3 tensor so covariant ⇔ curl E + ∂tB = 0 and div B = 0 (1.1.2) and (1.1.4) (1.3.15) while the two Maxwell inhomogeneous equations are (implied sum on μ ) ∂μFμν = μ0 Jν // both sides transform as a rank-1 tensor (4-vector) so covariant ⇔ curl B - μ0ε0∂tE = μ0J and div E = ρ/ε0 (1.1.24) and (1.1.15) (1.3.16) The fields are given by ( ε is the permutation tensor), B1 = -F23 E 1 = cF10 or B i = -(1/2)ε ijkFjk and Ei = cFi0 B2 = -F31 E2 = cF20 B3 = -F12 E 3 = cF30 . ( 1 . 3 . 1 7 ) The E and B fields are part of the tensor Fμν and so do not transform as four vectors like Aμ. That is to say, there are no 4-vectors of the form Eμ or Bν, so there is no up and down index on a field, so the index is just written down. Jackson states the above facts (but in Gaussian units) in his Section 11.9 along with a description of the noti on of covariance. Example : μ0J2 = ∂μFμ2 = ∂0F02 + ∂1F12 + ∂2F22 + ∂3F32 = (1/c)∂t(-1/cE2) + ∂1(-B3) + 0 + ∂ 3(+B1) = - (1/c2)∂tE2 + [curl B]2 => μ0J = curlB - μ0ε0∂tE in the 2 component Chapter 1: Basic Equations 34 (c) The Potential Wave Equations in th e King and Loren z Gauges with Conductors We refer to a certain gauge condition below as "the King gauge" because King (see Refs.) made extensive use of this condition in his books and papers at least as early as 1945. Perhaps this gauge has some official name, but we are not aware of it. We start with this King gauge and treat A and then φ. Then we do the Lorenz gauge case for A and φ, and finally we look at the wave equations for E and B. The motivation for using the King gauge is explained. Unlike most sources on this subject, we allow for the possibility that the conductors' μ i might differ from that of the dielectric. KING GAUGE Wave equation for A Consider the following general cross section of a tran smission line which happens to be of coaxial cable type, Fig 1.5 The gray regions 2 and 3 are conductors, while the white region 1 is the (possibly conducting) dielectric. Currents J1, J2 and J3 are conduction currents. We start by selecting the King gauge for region 1 and we apply it to all three regions, div A = - μ1ε1 ∂tφ - μ1σ1φ // ≡ King gauge, applied to all of R . (1.3.18) We first obtain the wave equation for A in region 1 . Start with (1.3.3) which gives the wave equation for A before any gauge choice is made, (∇2 - μ1ε1 ∂t2) A = grad [μ1ε1 ∂tφ + div A ] - μ1J . // region 1 (1.3.3) Now insert the King gauge (1.3.18) to get Chapter 1: Basic Equations 35 (∇2 - μ1ε1 ∂t2) A = grad [μ1ε1 ∂tφ + (- μ1ε1 ∂tφ - μ1σ1φ) ] - μ1J = - μ1σ1 grad φ - μ1J = - μ 1σ1 (-E -∂tA) - μ1(σ1E) . // from (1.3.1) and J = σ1E = - μ1σ1 ( -∂tA) . Thus the wave equation for A in region 1 is ( ∇ 2 - μ1ε1 ∂t2 - μ1σ1∂t) A = 0 // region 1 (1.3.19) This is a damped wave equation with no driving source; the equation is homogeneous. Now we start over with (1.3.3) for region 2: (∇ 2 - μ2ε2 ∂t2) A = grad [μ2ε2 ∂tφ + div A ] - μ2J2 // region 2 (1.3.3) As before, we insert the region-1 Ki ng gauge expression (1.3.18) for div A, even though we are now working in region 2, and we make an assumpti on that conductor 2 is a "very good conductor". ( ∇2 - μ2ε2 ∂t2) A = grad [μ2ε2 ∂tφ + (- μ1ε1 ∂tφ - μ1σ1φ) ] - μ2J2 = [ μ2ε2 ∂t + (- μ1ε1 ∂t - μ1σ1) ] gradφ - μ2J2 = [ ( μ 2ε2 - μ1ε1) ∂t - μ1σ1) ] gradφ - μ2J2 = [ ( μ 2ε2 - μ1ε1) ∂t - μ1σ1) ] (-E-∂tA) - μ2J2 // using (1.3.1) = [ ( μ2ε2 - μ1ε1) ∂t - μ1σ1) ] (-J2/σ2-∂tA) - μ2J2 // J2 = σ2E ≈ [ (μ2ε2 - μ1ε1) ∂t - μ1σ1) ] (-∂tA) - μ2J2 // since σ2 is very large in conductor 2 = - [ ( μ 2ε2 - μ1ε1) ∂t2 - μ1σ1∂t ] A - μ2J2 . The "large σ 2" assumption made two lines above is discussed at the end of this section. It puts a lower limit on the value ω for which the A wave equation is valid, but this limit is quite low relative to the normal use of a transmission line so it does not affect our analysis. Notice that we have chosen not to set J 2 = σ2E in region 2 for the last term, we just leave it as J2. Moving the first term on the right to the left we get Chapter 1: Basic Equations 36 ( ∇2 - μ2ε2 ∂t2 + [ (μ2ε2 - μ1ε1) ∂t2 - μ1σ1∂t ] ) A = - μ2J2 or ( ∇ 2 - μ1ε1 ∂t2 - μ1σ1∂t ) A = - μ2J2 . // region 2 (1.3.20) On the left side we see the same region-1 damped wave operator although we are in region 2, and J2 is the conduction current density in region 2. A similar result applies for region 3. Thus we have shown that ( ∇2 - μ1ε1 ∂t2 - μ1σ1∂t ) A = 0 region 1 ( ∇2 - μ1ε1 ∂t2 - μ1σ1∂t ) A = - μ2J2 region 2 ( ∇2 - μ1ε1 ∂t2 - μ1σ1∂t ) A = - μ3J3 region 3 (1.3.21) We combine these into a single equation which is then valid over all of region R, ( ∇2 - μ1ε1 ∂t2 - μ1σ1) A = - μ2J2 - μ3J3 all of region R (1.3.22) with the understanding that the conduction current in region 1 has already been accounted for and Ji represents conduction currents in conductor i .We co uld generalize this result for a region R containing any number N of conductors labeled i = 2,3...N+1 ( ∇2 - μ1ε1 ∂t2 - μ1σ1) A = - Σi=2N+1μiJi . all of region R (1.3.23) Wave equation for φ We first obtain the wave equation for φ in region 1. Start with (1.3.2) which gives the wave equation for φ before any gauge choice is made ∇2φ + ∂t[div A] = -ρ/ε1 . (1.3.2) Now use the same global region-R King gauge (1.3.18) for div A, ∇2φ + ∂t[- μ1ε1 ∂tφ - μ1σ1φ] = - ρ1/ε1 ( ∇2 - μ1ε1 ∂t2 - μ1σ1∂t)φ = - (1/ε1)ρ1 . // region 1 (1.3.24) Again the same damped region-1 wave operator appears on the left side. Since the King gauge is the same in all three regions, we can write ( ∇ 2 - μ1ε1 ∂t2 - μ1σ1∂t)φ = - (1/ε1)ρ(1) // region 1 ( ∇2 - μ1ε1 ∂t2 - μ1σ1∂t)φ = - (1/ε2)ρ(2) // region 2 ( ∇2 - μ1ε1 ∂t2 - μ1σ1∂t)φ = - (1/ε3)ρ(3) // region 3 (1.3.25) where ρ always means free charge. The three equations are basically the same because the pre-gauge equation (1.3.2) has no region-specific parameters apart from ε1, in contrast with (1.3.3) quoted above. Chapter 1: Basic Equations 37 Now the only actual free charge present is the surf ace charge on the outside surfaces of the conductors and we shall regard all these charge densities as residing in region 1, the dielectric (just inside the boundaries of region 1). Thus, write ρ (1) = ρ2 + ρ3 // = Σi=2N+1ρi ρ(2) = 0 ρ(3) = 0 ( 1 . 3 . 2 6 ) where ρ i is the surface charge density on conductor i. Then combine the above three equations into a single equation for all of region R ( ∇2 - μ1ε1 ∂t2 - μ1σ1∂t)φ = - (1/ε1) Σi=2N+1ρi . all of region R (1.3.27) Conclusion for wave equations in the King gauge Here then are the wave equations for φ and A in region R using the region-1 King gauge: Potential Wave Equations in the King Gauge (1.3.28) ( ∇2 - μ1ε1 ∂t2 - μ1σ1∂t)φ = - (1/ε 1) Σi=2N+1ρi all of region R (1.3.27) ( ∇2 - μ1ε1 ∂t2 - μ1σ1∂t)A = - Σi=2N+1 μiJi all of region R (1.3.23) div A = - μ1ε1 ∂tφ - μ1σ1φ King gauge (1.3.18) 1 = dielectric 2,3,4.... N+1= conduc tors (there are N conductors) ρi = free surface charge density on conductor i Ji = free current density in conductor i (J 1 in the dielectric exists but does not appear in ΣiμiJi) To be consistent with later sections, we put subscr ipt d on dielectric properties, and we renumber the conductors 1 to N instead of 2 to N+1. The above box then becomes Potential Wave Equations in the King Gauge (1.3.29) ( ∇2 - μdεd ∂t2 - μdσd∂t)φ = - (1/ε d) Σiρi all of region R ( ∇2 - μdεd ∂t2 - μdσd∂t)A = - ΣiμiJi all of region R div A = - μdεd ∂tφ - μdσdφ King gauge μd,εd,σd = dielectric 1,3,4.... N = conductors Σi = Σi=1N μi = for conductor i ρi = free surface charge density on conductor i Ji = free current density in conductor i (J in the dielectric exists but does not appear in ΣiμiJi) Chapter 1: Basic Equations 38 The word "free" is used above to emphasize the fact that possible polarization charge densities and magnetization current densities are not included in these ρi and Ji. King never writes these wave equations in his transmi ssion-line theory book, so it is difficult to find verification of our logic pathway in his book. However, Panofsky and Phillips do show the equations and we quote the relevant section from p 241 of their book: Their last sentence says that J = σE in the conducting dielectric has been incorporated into the -μσ∂ tA term in their first equation, just as we have done above. These authors have assumed that the μ's of the dielectric and the conductors are all the same (normally μ = μ0). In order to obtain the above equations, Panofsky and Phillips use the King gauge (1.3.18) but th ey refer to this gauge simply as "the Lorentz condition" (illustrating Comments 1 a nd 2 above). From their page 240, LORENZ GAUGE If we carry out the exact same program with respect to Fig 1.5 using a region-1 (dielectric) global Lorenz gauge for all of R, div A = - μ 1ε1∂tφ , ( 1 . 3 . 3 0 ) we obtain these results for A, where in region 1 the conduction current is not absorbed into a damping term on the left side, Chapter 1: Basic Equations 39 ( ∇2 - μ1ε1 ∂t2)A = - μ1J1 region 1 ( ∇2 - μ1ε1 ∂t2)A = - μ2J2 region 2 ( ∇2 - μ1ε1 ∂t2)A = - μ3J3 . region 3 As before, all three equations have the same wave operator on the left side. Again assuming N conductors, we combine these into a single equation as follows ( ∇ 2 - μ1ε1 ∂t2)A = - Σi=1N+1 μiJi all of region R (1.3.31) Meanwhile, the results for φ are ( ∇2 - μ1ε1 ∂t2)φ = -ρ(1)/ε1 region 1 ( ∇2 - μ1ε1 ∂t2)φ = -ρ(2)/ε2 region 2 ( ∇2 - μ1ε1 ∂t2)φ = -ρ(3)/ε3 region 3 so that with the same comments made earlier in (1.3.26) this becomes ( ∇2 - μ1ε1 ∂t2)φ = - (1/ε 1) Σi=2N+1 ρi all of region R (1.3.32) We then make the same notational change ma de above to get these Lorenz-gauge results: ( ∇2 - μdεd ∂t2)φ = - (1/ε ) Σi=1N ρi all of region R (1.3.33) ( ∇2 - μdεd ∂t2)A = - Σi=1N μiJi - μJ all of region R (1.3.34) Notice that no conductivities appear in these equations. COMPARISON We can now do a side-by-side comparison, where Σ i is a sum over the conductors i = 1,2..N King Gauge: ( ∇2 - μdεd ∂t2 - μdσd∂t)φ = - (1/ε d) Σiρi all of region R (1.3.29) ( ∇2 - μdεd ∂t2 - μdd∂t)A = - ΣiμiJi all of region R (1.3.29) div A = - μdεd ∂tφ - μdσdφ King gauge (1.3.29) Lorenz Gauge: ( ∇2 - μdεd ∂t2)φ = -(1/εd) Σiρi all of region R (1.3.33) ( ∇2 - μdεd ∂t2)A = - Σi μiJi - μdJ all of region R (1.3.34) div A = - μdεd ∂tφ Lorenz gauge (1.3.30) In the Lorenz gauge, we get undamped wave operators, but the sum on the right of the A equation includes the current J in the dielectric, whereas this is not the case in the King gauge. In a situation where we have prescribed currents Ji in the conductors, it is inconvenient to have to worry about the dielectric conduction current J which complicates the solution of the problem . In the King gauge, we get damped Chapter 1: Basic Equations 40 wave operators but we have to include only the current in the conductors since the current in the dielectric has been incorporated into the damping term. When we transform to the freque ncy domain and write the Helmholtz equation for A and its Helmholtz Integral solution, we need only integrate over the conductors which makes life easier. This then is the motivati on for the King gauge. For a non-conducting dielectric both gauge conditions are the same since σ = 0. E AND B WAVE EQUATIONS Meanwhile, the E and B field wave equations of course don't know anything about gauges and from (1.2.1) and (1.2.2) we have, with respect to Fig 1.5, ( ρ s = ρ2+ρ3 = Σi=2N+1ρi, 1 = dielectric) ( ∇2 - μ1ε1 ∂t2)E = μ1∂tJ1 + (1/ε1) grad ρ(1) = μ1∂tJ1 + (1/ε1) grad ρs // region 1 ( ∇2 - μ2ε2 ∂t2)E = μ2∂tJ2 + (1/ε2) grad ρ(2) = μ2∂tJ2 // region 2 ( ∇2 - μ3ε3 ∂t2)E = μ3∂tJ3 + (1/ε3) grad ρ(3) = μ3∂tJ3 // region 3 ( 1 . 3 . 3 5 ) ( ∇2 - μ1ε1 ∂t2)B = - μ1 curl J1 // region 1 ( ∇2 - μ2ε2 ∂t2)B = - μ2 curl J2 // region 2 ( ∇2 - μ3ε3 ∂t2)B = - μ3 curl J3 // region 3 where Ji = σiE . We cannot unify each group of three equa tions into a single region R equation as we could in the potential case since the wave ope rators are different in each region. Using Ji = σiE and curl E = - ∂tB and (1.3.26) the above equations can be rewritten as, ( ∇2 - μ1ε1 ∂t2 - μ1σ1∂t)E = (1/ ε1) Σi=2N+1 grad ρi // region 1 ( ∇2 - μ2ε2 ∂t2 - μ2σ2∂t)E = 0 // region 2 ( ∇2 - μ3ε3 ∂t2 - μ3σ3∂t)E = 0 // region 3 ( 1 . 3 . 3 6 ) ( ∇ 2 - μ1ε1 ∂t2 - μ1σ1∂t)B = 0 // region 1 ( ∇2 - μ2ε2 ∂t2 - μ2σ2∂t)B = 0 // region 2 ( ∇2 - μ3ε3 ∂t2 - μ3σ3∂t)B = 0 // region 3 Again the three damped wave operators are different. The solution of these equations requires solving the first equation for the particular solution in region 1, finding all possible homogenous solutions to all 6 equations in their regions using appropriate harmonic forms with "cons tants to be determined", then matching these conditions at the two boundaries to evaluate the constants. In contrast, in the potential problem of (1.3.28), ( ∇ 2 - μ1ε1 ∂t2 - μ1σ1∂t)φ = - (1/ε 1) Σi=2N+1ρi all of region R (1.3.26) ( ∇2 - μ1ε1 ∂t2 - μ1σ1∂t)A = - Σi=2N μiJi all of region R (1.3.23) div A = - μ1ε1 ∂tφ - μ1σ1φ King gauge (1.3.18) (1.3.28) one worries about a single unified region R and there is only one damped wave operator. The method of solution is to find the particular solutions of the φ and A equations, add in ho mogenous solutions and match boundary conditions. Chapter 1: Basic Equations 41 The "large σ2" assumption . This assumption was used above in the development of the region 2 damped wave equation (1.3.20) for A . Looking back at the development one sees that the approximation made was in fact | E| << |∂ tA| inside the conductor. An estimation of the validity of this inequality requires material that appears in later chapters, so we assume that material in what follows. It will turn out that we only care about the z component of the A wave equation which involves A z, because the transverse components of A are so small that they can be neglected (Appendix M and self-c onsistency). Thus, we want to show that |E z| << |ω Az| where E z and Az are now in the frequency domain. From the study of "the transverse problem" in Chapter 5, we can make a ballpark estimate that A zt(x,y) ~ K, where A zt is a certain transverse version of A z, and where K is a certain dimensionless constant arising in the theory. This estimate for A zt arises from the boundary conditions on A zt shown in (5.3.11). The connection between A z and Azt is given in (5.2.1) Az(x,y,z) = μd 4π i(z) Azt( x , y ) (5.2.1) so we then have A z ~ (μd/4π) i(z) K = ( μd/4π) I K where i(z) = I is the current in a conductor. For a round wire of radius a we can estimate J z = I/(πa2). Then from J z = σEz we have E z ~ I/(π a2σ). The inequality in question is then |E z | << |ωAz| I/(π a 2σ) << ω (μd/4π) I K 1/(a 2σ) << ω (μdK/4) ω >> 1 a2 4 σμK or f >> 1 a 2 2 π 1 K 1 μσ . ( 1 . 3 . 3 7 ) For a copper conductor, μσ ≈ 4π * 5.81 = 73.0 sec/m2 so then f >> 1 a2 2 π 1 K 1 73 = .0087/ (a2K) . (1.3.38) In order to justify our "large σ2" assumption, we require that the operating frequency be significantly larger than .0087/ (a2K). We will show in the following two ex amples that this is quite a low frequency and one always operates above this lower limit in a practical application. Example 1: Belden 8281 coaxial cable is treated as a case study in Appendix R. For the central conductor, a = 394 μ and the the cable has K = 3.7. Our condition is then f >> 15 KHz, Chapter 1: Basic Equations 42 Since 15 KHz is an audio frequency, while Belden 8281 coaxial cable is used for RF signals, this lower limit is not an issue. That is to say, the A z wave equation (1.3.20) is valid for ω of practical use. Example 2: At the end of Section 4.6 below we consid er a power distribution transmission line which has two conductors with a = 1/2" and K = 17.5. For such a transmission line, our condition is f >> 3 Hz, Since power systems operate at 50 or 60 Hz , this lower bound of 3 Hz is well surpassed. Chapter 1: Basic Equations 43 1.4 Retarded Solutions in the Lorenz gauge: Propagators In a mediu m where μ and ε are time-independent, the Lorenz gauge equations (1.3.4) and (1.3.5) apply, ( ∇2 - με ∂t2)φ = - (1/ε )ρ (1.3.4) ( ∇2 - με ∂t2)A = - μJ . (1.3.5) One approach to solving these equations for A and φ is the method of retarded solutions. We seek to solve an equation of this form ( ∇2 - με ∂t2) u = - f , (1.4.1) where for example in (1.3.4) u = φ and f = ρ /ε. Since με = 1/v2 where v is the wave velocity in the medium, write (1.4.1) as (∂t2 - v2∇2) u = v2f or … u = f where … ≡ 1 v2 ∂t2 - ∇2 . ( 1 . 4 . 2 ) This last equation is similar to (A .7.2) of Appendix A and can be solv ed in the same manner. Define a Green's function g as the solution of v 2 … g(x,t; x',t') = δ (x-x')δ(t-t') with g = 0 when | x-x'|→∞ . (1.4.3) This is just (A.7.3) with c = v. As (A .7.4) shows, the solution is given by v2 g(x,t; x',t') = (1/4 πR)δ(t-t'-R/v) with R = | x-x'| . (1.4.4) The delta function only gets a hit if t = t'+R/v, so there is never a hit if t < t'. In other words, g = 0 for t<t', and g is often referred to as a "causal" Green 's function. Jackson (6.41) and (6.44) uses G(+) = 4πv2g with v = c and refers to the solution as a "retarde d Green function". See also Stakgold references in Appendix A. The solution to (1.4.1) is then u( x,t) = ∫d3x' ∫dt' v2 g(x,t; x',t') f(x ' , t ' ) ( 1 . 4 . 5 ) as can be verified by applying … to both sides and making use of (1.4.3). The Green's Function g( x,t; x',t') is the free-space fundamental solution (propagator) of the wave equation. Insert (1.4.4) into (1.4.5) to get, u( x,t) = ∫d3x' ∫dt' (1/4πR)δ(t-t'-R/v) f( x',t') = ∫d3x' (1/4πR) f( x', t-R/v) = 1 4π ∫d3x' f(x',t-R/v) R . ( 1 . 4 . 6 ) Chapter 1: Basic Equations 44 Thus, the solutions to (1.3.2) and (1.3.3) are (Lorenz gauge) : φ(x,t) = 1 4πε ∫d3x' ρ(x',t-R/v) R ( 1 . 4 . 7 ) A(x,t) = μ 4π ∫d3x' J(x',t-R/v) R . R = |x-x'| (1.4.8) The potentials at time t are generated by the values the sources had at time t - R/v since the influence of the sources travels at finite velocity v through the me dium. These last equations agree with Jackson p 246 (6.48). Note that 1/4π R is the free-space propagator of the Poi sson equation. It describes how a source at location x' and earlier time t-R/v propagates its influe nce into the potential at observation point x and current time t. Compare (1.4.6) to (A.0.2) which is the solution to the electro static Poisson equation, where the source has no time dependence (it is static). Jumping the gun slightly, it is interesting now to Fourier Transform the above equations. First, write ρ(x',t-R/v) = ∫-∞ ∞ dt' δ(t'-[t-R/v]) ρ (x',t') . (1.4.9) Then using the Fourier Integral Transform (1.6.8), φ(x,ω) = ∫-∞ ∞ dt φ(x,t)e-jωt / / ( 1 . 6 . 8 a ) = ∫-∞ ∞ dt [1 4πε ∫d3x' ρ(x',t-R/v) R ] e-jωt // insert φ from (1.4.7) = 1 4πε ∫-∞ ∞ dt ∫d3x' 1 R ∫-∞ ∞ dt' δ(t'-[t-R/v]) ρ(x',t') e-jωt // insert ρ(x',t-R/v) from (1.4.9) = 1 4πε ∫d3x' 1 R ∫-∞ ∞ dt' ρ(x',t') e-jω[t'+R/v] // do the dt integration = 1 4πε ∫d3x' e-jβR R ∫-∞ ∞ dt' ρ(x',t') e-jωt' // let β ≡ ω/v = 1 4πε ∫d3x' e-jβR R ρ(x',ω) . // (1.6.8a) Thus, in the frequency domain the re tarded potential solutions appear as Chapter 1: Basic Equations 45 φ(x,ω) = 1 4πε ∫d3x' e-jβR R ρ(x',ω) ( 1 . 4 . 1 0 ) A(x,ω) = μ 4π ∫d3x' e-jβR R J(x',ω) . R = |x -x'| β = ω/v (1.4.11) These are the single-region expressions of the Helmholtz integrals we shall obtain in the next section by a somewhat different path using a different gauge. These integrals then are the ω-domain versions of the retarded potential solutions in the time domain. The factor e -jβR/R is the ω-space 3D Helmholtz propagator discussed below and in Appendix H. It describes how the ω-domain source ( ρ or J) at location x' propagates to its potential at location x. These last two equations have the general form f 1(x) = ∫k(x,x')f2(x')d3x ' ( 1 . 4 . 1 2 ) and the propagator k( x,x') is sometimes called "the kernel" and de fines an integral operator K. Then the above equation is written f 1 = Kf2 which is a mapping from one function to another in a Hilbert Space of functions. Similarly, equation (1.4.5) has the form f 1(x,t) = ∫∫k(x,t; x',t') f2(x',t') d3x' dt' (1.4.13) where now the kernel k(x ,t; x',t') is a spacetime propagator describing how f 2 at x' and t' contributes to f 1 at x and t. The total function f 1 is the sum of all these propagated contributions. For the particular propagator shown in (1.4.5), f 1(x,t) would only get contributions from f 2(x',t') at past times t', so that k is a causal propagator. The same notion of f 1 = Kf2 applies. Comment: The word "propagator" is commonly used in quantum mechanics where the entity being propagated is a probability amplitude , and the total amplitude for some "event" is the sum of all the propagated contributions. This viewpoint was pr omoted by Richard Feynman, and the graphical representation of equations like (1.4.12) is called a Feynman Diagram : Fig 1.6 Chapter 1: Basic Equations 46 1.5 The Wave Equations in the Frequency Domain (a) The Transformed Wave Equations A standard method of solving wave equations invo lves transforming the equations from the time domain to the frequency ω domain using the Fourier Integral Transform, assuming that the μ, ε and σ are constants (possibly complex). As an example, we start with the φ equation in (1.3.29) and expand φ(x,t) and ρs(x,t) onto their Fourier components using (1.6.8). The overloaded notation is explained in Section 1.6 (f). ( ∇ 2 - μdεd ∂t2 - μdσd∂t) φ(x,t) = - (1/εd) Σiρi(x,t) (1.3.26) ( ∇2 - μdεd ∂t2 - μdσd∂t) [(1/2π) ∫-∞ ∞ dω e+jωt φ(x,ω)] = - (1/εd) [(1/2π) ∫-∞ ∞ dω e+jωt Σiρi(x,ω) ] ∫-∞ ∞ dω (∇2 - μdεd ∂t2 - μdσd∂t) e+jωt φ(x,ω) = - (1/εd) ∫-∞ ∞ dω e+jωt Σiρi(x,ω) ∫-∞ ∞ dω (∇2 + μdεdω2 -jω μdσd) e+jωt φ(x,ω) = - (1/εd) ∫-∞ ∞ dω e+jωt Σiρi(x,ω) ∫-∞ ∞ dω e+jωt [(∇2 + μdεdω2 - jω μdσd) φ(x,ω)] = ∫-∞ ∞ dω e+jωt [- (1/εd) Σiρi(x,ω)] . At this point we invoke the completeness of the set of functions {ejωt} on the interval (- ∞,∞) to claim that the integrands must be equal, giving (1.3.26) transformed to the frequency domain, ( ∇2 + μdεdω2 - jω μdσd) φ(x,ω) = - (1/εd) Σiρi(x,ω) . or ( ∇2 + βd2) φ(x,ω) = - (1/ε d) Σiρi(x,ω) where β d2 is the following complex "Helmholtz parameter" [of Helmholtz operator ( ∇2 + βd2) ], βd2 = μdεdω2 - jωμdσd = ω2μd ( εd - jσd/ω) = ω2μdξd ξd ≡ εd - jσd/ω . (1.5.1a) β d02 = ω2μdεd when σd = 0 (non-conducting dielectric) (1.5.1b) Here ξ d(ω) is the "complex dielectric constant", nothing more or less than the expression shown. We shall have occasion (mainly in Appendix D) to use the damped wave equation for the E field inside a transmission line conductor. We referred to such conductors as region 2 or region 3 in the discussion above, but here we shall use no subscript to denote pa rameters inside a conductor. Looking at (1.3.36), such a wave equation when converted to the frequency domain becomes, ( ∇ 2 + β2) E(x,ω) = 0 (1.5.27) Chapter 1: Basic Equations 47 where β 2 = μεω2 - jωμσ = ω2μ ( ε - jσ/ω) = ω2μ ξ ξ ≡ ε - jσ/ω . (1.5.1c) and all parameters here refer to th e conductor. This has the same form as (1.5.1a) but with no dielectric subscripts. For copper, we show later in (2.2.3) that for f << 1018 Hz, one can neglect the ε term in the above expression for β2 which then gives β2 = - jωμσ ( 1 . 5 . 1 d ) Note: Hermann von Helmholtz (1821-1894) was an early electromagnetic researcher and equations of the form (∇2+k2)f = g bear his name. As we have just seen, his equation arises from a temporal Fourier or Laplace transform of a wave equation. Since k will have another meaning in Chapter 5, and to be consistent with King p 10 (15a,b,c), we define the quantities in (1.5.1) as β2 instead of k2. King bolds parameters when they are complex, but we do not, so we have β2 instead of β2. Examination of the above transformation shows that any equation can be tran sformed from the time domain to the frequency domain using these simple rules, ∂ t → +jω ∂t2 → -ω2 F( x,t) → F( x,ω) . (1.5.2) where it is understood (Section 1.6) that F( x,t) and F( x,ω) are different functions. Thus, the frequency-domain representations of the King-gauge potential wave equations shown in (1.3.29) are: Potential Wave Equations in the King Gauge (ω domain) ( ∇2 + βd2)φ = - (1/εd) Σiρi all of region R (1.5.3) ( ∇2 + βd2)A = - Σi μiJi all of region R (1.5.4) div A = - μdεdjωφ - μdσdφ = -jωμd(εd+σd/jω)φ = -jωμdξdφ = -j(βd2/ω)φ King gauge (1.5.5) βd2 = μdεdω2 - jωμdσd = ω2μd ( εd - jσd/ω) = ω2μd ξd ξd ≡ εd - jσd/ω (1.5.1a) μd,εd,σd = dielectric 1,3,4.... N = conductors Σi = Σi=1N μi = for conductor i ρi = free surface charge density on conductor i Ji = free current density in conductor i (J in the dielectric exists but does not appear in ΣiμiJi) In these equations, all mathematical fields φ, A, ρi, Ji are functions of x and ω. Note from (1.5.5) that in the ω domain, the King gauge is the Lorenz gauge with εd → ξd. Chapter 1: Basic Equations 48 (b) The Helmholtz Integrals in the King Gauge The next step is to solve the above equations for φ and A. The method was demonstrated in Appendix A.0 and is applied again here. We first define the fr ee-space Green's Function g by this boundary value problem [ back to a generic β parameter ] - (∇2 + βd2)g(x,x') = δ (x-x') where lim |x|→∞ g(x,x') = 0 . (1.5.6) As shown in (H.1.5), the solution to problem (1.5.6) is g( x,x') = 1 4π e-jβdR R R = | x - x'| . (1.5.7) As (1.5.1a) shows, in a conducting dielectric βd2 has a small negative phase, so βd has half this negative phase and β d then has a small negative imaginary part. Then e-jβdR → 0 for large R, as required by the condition of problem (1.5.6). This is why e+jβdR /R is a rejected solution. The Helmholtz equations (1.5.3) and (1.5.4) have the following particular solutions (dV' = d3x'), φ(x,ω) = 1 4πε Σi∫ρi(x',ω) e-jβdR R dV' R = | x - x'| (1.5.8) A(x,ω) = 1 4π Σi∫μiJi(x',ω) e-jβdR R dV' R = | x - x'| . (1.5.9) In these equations, βd is a function of ω, namely βd = ω2μdξd as in (1.5.1a), and Σi = Σi=1N is over the conductors. Since ( ∇2 + βd2) is the Helmholtz operator, solutions of the form (1.5.8) and (1.5.9) are sometimes called "Helmholtz integrals". To verify that the φ of (1.5.8) solves (1.5.3) we write φ(x,ω) = ∫ [Σiρi(x',ω)/ε] g(x,x') dV' so that, - (∇2 + βd2) φ(x,ω) = ∫[ Σiρi(x',ω)/ε] { - ( ∇2 + βd2)g(x,x') } dV' = ∫[ Σiρi(x',ω)/ε] {δ(x-x')} d3x' = Σiρi(x,ω)/ε . In the limit ω→ 0 we find from (1.5.1a) that βd(ω) → 0 and then (1.5.9) is th e same as (A.0.2) obtained from electrostatics and Poisson's Equation. Chapter 1: Basic Equations 49 In (1.5.8) the volume density function ρs(x',ω) ≡ Σiρi(x',ω) represents a surface charge density, so it is convenient to represent φ as a surface integral over the corresponding surface charge density n s(x',ω), φ(x,ω) = 1 4πε ∫ns(x',ω) e-jβdR R dS' . R = | x - x'| (1.5.10) Comments on n, σ and Dirichlet : Usually one uses σ for a surface charge, but σ is already used for conductivity so we use n. To further complicate things, in his potential theory discussion of Chapter 6, Stakgold uses σ to represent our surface S enclosing a volume V (his region R) as in our Fig 1.2. Stakgold uses n to indicate a normal derivative, as in this Dirichlet problem solution of the Poisson equation - ∇2φ(x) = q(x), φ(x) = ∫R dx' g(x|x') q(x') – ∫σ dSξ f(ξ) ∂ξng(x|ξ) // Stakgold (6.81) . (1.5.11) Here ∂ξn = ∂/∂nξ where n ξ is a local coordinate on the surface σ at point ξ which is normal to the surface. In this equation, q(x) is the Poisson source (think ρ (x)/ε0), g(x|ξ) is the full Green's function, meaning g = 0 on boundary σ, and f(ξ) is the Dirichlet prescribed potential on the enclosing boundary σ. Stakgold also uses n for number of dimensions and his work is always done in n spatial dimensions. In (1.5.11), the first term is the particular solu tion, like our Helmholtz inte gral, while the second term is a homogenous solution to - ∇2φ(x) = 0 which, when added in, ma kes things work at boundaries. Our Helmholtz integral, however, uses the free-space Green's Function, so we cannot just add on Stakgold's Dirichlet term to get a solution. The above Poisson Dirichlet solution (1.5.11) seems my sterious at first viewing, but is easily derived using - ∇2g(x|x') = δ(x-x') , -∇2φ(x) = q(x) [ = ρ(x)/ε ], and the famous Green's 2nd "symmetric" identity, where ∂φ/∂n = n^ • ∇φ = the same normal derivative ∂ξn discussed above, ∫V dV ψ ∇2φ = ∫S dS ψ( ∂φ/∂n) – ∫V dV (∇ψ • ∇φ) Green #1 ∫V dV [ ψ ∇2φ – φ ∇2ψ ] = ∫S dS [ ψ (∂φ/∂n) – φ (∂ψ/∂n) ] . Green #2 (1.5.12) Here #2 = #1( ψ,φ) - #1(φ ,ψ) and #1 is derived from the divergence theorem (1.1.30) with F = ψ∇φ and vector identity ∇• (ψ∇φ) = ∇ψ • ∇φ + ψ ∇2φ and d S = dS n^ . Green was a busy man. Equation (1.5.11) is then obtained by setting ψ = g in (1.5.12), recalling that g = 0 on σ. Since Green #2 is also valid if we replace ∇2→ (∇2+k2), (1.5.11) is also formally valid for a Helmholtz Dirichlet problem where then g is the full Helmholtz Green's function. (c) King's leading factor (1/4 πξ) and t he final Helmholtz Integrals This is a somewhat subtle point and something th at King never discusses much in his transmission-line theory book. The issue is that there are two different entities n s and nc which have units charge/area, and they are related by n s = (εd/ξd) nc where ξd = εd + σd/jω is the complex dielectric constant (in the dielectric) which incorporates the effect of possible dielectric conductivity. In a transmission line Chapter 1: Basic Equations 50 problem, it is n c that is specified by the boundary conditions and not n s (which is the actual surface charge density). For that reas on, one replaces (1.5.10) with, φ(x,ω) = 1 4πξd ∫nc(x',ω) e-jβdR R dS' R = | x - x'| (1.5.13) which explains the leading factor 1 4πξ which appears every time King writes down the Helmholtz integral for φ in his books. In the discussion below we describe n c and its relation to n s, and then we show how this relation works in the simple example of a parallel plate capacitor. Consider the situation at a general boundary between dielectric (region 1) and conductor (region 2) where there exists a surface charge density n s : Fig 1.7 In (1.1.18) it was shown that div [ Jd + Jc] = 0 where J c = σ E is the conduction current and Jd the displacement current ∂ tD = ε∂tE. The divergence theorem (1.1.30) then says 0 = ∫V div [Jd + Jc] dV = ∫S [Jd + Jc] • dS . Applied to the blue pillbox which stra ddles the boundary in the figure, we find Jd1n + Jc1n = Jd2n + Jc2n where n means normal component. Writing this out, ε 1∂tE1n + σ1E1n = ε2∂tE2n + σ2E2n ≈ σ2E2n = Jc2n since σ2 is huge inside the conductor. Therefore, Chapter 1: Basic Equations 51 Jc2n = σ1E1n + ε1∂tE1n . ( 1 . 5 . 1 4 ) Meanwhile, Gauss's Law (1.1.33) states that div (εE) = ρ ⇔ ∫V ρ dV = ∫S εE • dS . (1.1.33) Applied to the same blue pillbox we find n s = ε1En1 - ε2En2 ≈ ε1En1 since E n2 ≈ 0 inside the conductor. Thus, E n1 = ns/ε1 and then J cn1 = σ1En1 = ns(σ1/ε1) . (1.5.15) Then (1.5.14) can be written as J c2n = σ1E1n + ε1∂tE1n = (σ1 + ε1∂t)E1n = (1/ε1)(σ1 + ε1∂t)ns or, writing out the arguments, Jc2n(x,t) = (1/ε1)(σ1 + ε1∂t)ns(x,t) . In the frequency domain with rules (1.5.2) this becomes Jc2n(x,ω) = (1/ε1)(σ1 + ε1jω)ns(x,ω) = ( 1 / ε 1) (jω)(ε1 + σ1/jω) ns(x,ω) = ( ξ 1/ε1) (jω) ns(x,ω) . ξ1 ≡ ε1 + σ1/jω = complex dielectric constant (1.5.16) If we observe the conduction current J c2n flowing through a unit-area loop (red in figure), we can write Jc2n = ∂tnc where n c is the total amount of conduction charge flowing through that unit-area loop per unit time. Thus we have ∂ tnc(x,t) = (1/ε1)(σ1 + ε1∂t)ns(x,t) or jω nc(x,ω) = (ξ1/ε1) (jω) ns(x,ω) or n c(x,ω) = (ξ1/ε1) ns(x,ω) . ( 1 . 5 . 1 7 ) where is our result claimed at the start that n s = (εd/ξd) nc. Note that: • The quantity n s is the amount of free charge per unit area on the conductor surface. Chapter 1: Basic Equations 52 • The quantity n c does not represent any kind of surface charge anywhere (free or otherwise). nc is related to the transport of conduction charge carriers through the charge-neutral interior of the conductor just below the surface. There is no unit-area surface which holds n c amount of charge, but both ns and nc have the dimensions of charge/area so both can therefore be called "surface charge". These two areal charge densities are different simply because the dielectric leaks charge off the surface. We are now going to rederive (1.5.17) a different way. We can write, using the blue pillbox and continuity relation (1.1.25), div Jc = - ∂tρfree ⇔ -∂t[∫V ρfree dV] = ∫S Jc • dS . (1.1.25) => - ∂t[∫V ns dS] = ∫S Jc • dS => -∂tns = Jcn1 - Jcn2 = σ1(ns/ε1) - ∂tnc // see above: J cn1 = σ1(ns/ε1), Jcn2 = ∂tnc so ∂tns = ∂tnc - σ1(ns/ε1) // change in n s = flow in - flow out or jωn s = jω nc - σ1ns/ε1 => (j ω+ σ1/ε1)ns = jωnc => (jωε1+ σ1)ns = jωε1nc => (ε 1+σ1/jω)ns = ε1nc => ξ1 ns = ε1 nc => n c = (ξ1/ε1)ns which is the same as (1.5.17). Note that surface charge n s is real, while n c is complex. It is useful at this point to examine the simple cas e of a parallel plate capacitor to see the meaning of n s and nc. The plate separation s is meant to be very sma ll compared to the transverse dimensions of the plates, so the picture is distorted. We drop the subscript 1 on dielectric properties. Fig 1.8 Chapter 1: Basic Equations 53 First off, a DC analysis of the above devi ce shows that the capacitor has resistance R, R = V I = V JA = Es σdEA = (s/σdA) . (1.5.18) Now we assume an AC voltage V. The total cu rrent entering the conducting capacitor is I = J cA. If we think of I = ∂tQ then Q is the amount of charge passing through the external wire per unit time. Q is not the total charge on the left plate surface which in fact is Q s = nsA. Since I = J cA we have ∂tQ = (∂tnc) A and therefore Q = n cA. Meanwhile, the voltage V between the plates is V = Es, and we know that E = n s/εd from Gauss's law. Thus V = (s/ εd)ns. If we define the (complex) capacitance by Q = C'V, then C' = Q V = ncA (s/εd)ns = nc ns (Aεd/s) = (ξd/εd) (Aεd/s) = (ξd/εd) C = (Aξd/s) . (1.5.19) The capacitance C' is complex because it accounts for both the capacitance and conductance of the dielectric, C' = ε d + σd/jω εd (Aεd/s) = (Aεd/s) + (σdA/s)/(jω ) = C + 1/(j ωR) (1.5.20) or jωC' = jωC + 1/R or 1 Z = 1 Xc + 1 R Z = X c' = 1 jωC' X c = 1 jωC (1.5.21) which is the rule for computing an impedan ce Z for a capacitor and resistor in parallel Fig 1.9 Looking back at this example, it is clear that if one wants to compute the complete impedance of the conducting capacitor, one uses C' = Q/V where Q = An c. The ratio Q s/V gives only the capacitance C. Qs V = nsA (s/εd)ns = (Aεd/s) = C . (1.5.22) In this conducting capacitor problem, the boundary conditions are the voltage V or the total current I. Specification of the current I = ∂t(ncA) is really a specification of n c since in the frequency domain we Chapter 1: Basic Equations 54 then have I = j ωAnc. In analyzing the problem in full, we are thus interested in working with n c and not ns. So recalling now the King gauge Helmholtz integral for φ , φ(x,ω) = 1 4πε ∫ns(x',ω) e-jβdR R dS' R = | x - x'| , (1.5.10) since it will be more convenient to have n c in the integrand, we use (1.5.17) that n s = (εd/ξ(d) nc to rewrite the above expression as φ(x,ω) = 1 4πξd ∫nc(x',ω) e-jβdR R dS' R = | x - x'| (1.5.13) which is just (1.5.13) stated earlier. So here are our final forms of the Helmholtz integrals of interest, where now write n c = Σinci, φ(x,ω) = 1 4πξd Σi∫nci(x',ω) e-jβdR R dS' R = | x - x'| (1.5.13) A(x,ω) = 1 4π Σi∫μiJi(x',ω) e-jβdR R dV' R = | x - x'| (1.5.9) β d2 = μdεdω2 - jωμdσd = ω2μd ( εd - jσd/ω) = ω2μd ξd ξd ≡ εd - jσd/ω (1.5.1) where the sum Σi is over all conductors. If a ll conductor have the same μi = μc, (1.5.9) simplifies to A(x,ω) = μc 4π Σi∫Ji(x',ω) e-jβdR R dV' R = | x - x'| . (1.5.9)' We now quote directly from King's Transmission-Line Theory book to show how he presents the Helmholtz integrals for φ and A. What we call the King gauge appears as (2b) below. His symbols σ , ε, μ, ξ and β apply to the dielectric. // page 8 // page 9 // page 11 Chapter 1: Basic Equations 55 Comments: (1) King's (23) and (24) are for one conductor, while our (1.5.13) and (1.5.9) are for several conductors. (2) Due to time lag effects, ε and σ may be complex, so ε = ε'-jε" and σ = σ'-jσ". In this case ξ d ≡ εd - jσd/ω = (ε'd-jε"d) - j(σ 'd-jσ"d)/ω = [ε'd- σ"d/ω] - j [σ 'd + ωε"d]/ω = εeff - jσeff/ω so one would replace ε d → εeff and σd → σeff in all equations (see King p 9 footnote). (3) In the same way, time lag effects can cause μ = μ' - jμ" (hysteresis) [ generic μ ] (4) King uses bold font for vectors and for quantities which are complex. For example his ξ of ξ = ε - jσ/ω is bolded. Similarly, our (1.5.1) that β 2 = ω2μξ becomes his equation (10) above, β2 = ω2μξ . He does not use d subscripts on dielectric parameters as we do. (5) King assumes that all conductors and the dielectric have the same μ, something we did not assume. In order to make (23) and (24) look as similar as possible, he defines ν ≡ 1/μ. Since these parameters can both be complex, he writes them as μ and ν. This then explains the factor 1/(4 πν) appearing in his (24) which then agrees with our (1.5.9)'. (6) He shows his equation (23) charge density n' in bold, indicating it is complex. His n' is our n c, also complex. He refers to n ' as "charge density on the surface" but he really means it to be n c as we have discussed at length above, and this is how he uses it in his calculations. King uses these Helmholtz integrals (23) and (24) for φ and A extensively in his book to compute the parameters of various complicated transmission line geometries and interfaces. We shall pursue this subject more in Chapter 4 for some simple cases. We should point out that King makes no attempt to deri ve his equations (23) and (24) and more or less just pulls them out of a hat. We spent some time perusing several of King's other 11 books looking for some kind of derivation but were unsuccessful. The equations do appear in more or less the same form in his earliest book Electromagnetic Engineering (1945). So in some sense, we have spent the first 40 pages of this Chapter deriving his equations (23) and (24). Fo r that reason, it is worth gathering up the results in a summary box: Chapter 1: Basic Equations 56 Potential Solutions for φ and A in the King Gauge ( ω space) (1.5.23) φ(x,ω) = 1 4πξd Σi∫nci(x',ω) e-jβdR R dS' R = | x - x'| (1.5.13) φ(x,ω) = 1 4πξd Σi∫ρci(x',ω) e-jβdR R dV' // using volume charge representation ρcdV' = ncdS' A(x,ω) = 1 4π Σi∫μiJi(x',ω) e-jβdR R d V ' (1.5.9) A (x,ω) = μ 4π Σi∫Ji(x',ω)] e-jβdR R dV' // if all μi = μ (1.5.9)' μd,εd,σd = dielectric; μi = inside conductor i ; βd2 = μdεdω2 - jωμdσd = ω2μd ( εd - jσd/ω) = ω2μd ξd ξd ≡ εd - jσd/ω (1.5.1a) div A = - μdεdjωφ - μdσdφ = -jωμd(εd+σd/jω)φ = -jωμdξdφ // King gauge (1.5.5) B = curl A E = - grad φ - ∂tA (1.3.1) The Helmholtz integrals are just "particular solutions" to the potential wave equations. In order to solve a problem, one must add to these particular soluti ons whatever homogeneous solutions are necessary in order to match all boundary conditions. Chapter 1: Basic Equations 57 (d) Frequency domain wave equations for fields and potentials i n the Lorenz Gauge We now use the earlier notation with reference to Fig 1.5 where dielectric = 1 and conductors = 2,3...N+1 for N conductors. The Lorenz gauge is gi ven by (1.3.30) transformed to the ω domain, div A = - μ 1ε1jωφ . ( 1 . 5 . 2 4 ) Undamped Lorenz-gauge potential wave equations (1.3.32) and (1.3.31) : k 12 = ω2μ1ε1 ( ∇2+k12)φ = - (1/ε 1) Σi=2N+1 ρi all of region R ( ∇2+k12)A = - Σi=1N+1 μiJi all of region R (1.5.25) In the Lorenz gauge, the potential wave equations don't have damped ope rator versions. However, for the field wave equations (which know nothing of gauge) we can write bot h undamped and damped versions: Undamped field wave equations (1.3.35) : k i2 = ω2μiεi ( ∇2+k12)E = μ1jωJ1 + (1/ε1) Σi=2N+1 grad ρi (∇2+k12)B = - μ1 curl J1 // region 1 ( ∇2+k22)E = μ2jωJ2 ( ∇2+k22)B = - μ2 curl J2 // region 2 ( ∇2+k32)E = μ3jωJ3 ( ∇2+k32)B = - μ3 curl J3 // region 3 (1.5.26) Damped field wave equations (1.3.36) : β i2 = ω2μiξi ( ∇2+β12)E = (1/ε1) Σi=2N+1grad ρi ( ∇2+β12)B = 0 // region 1 ( ∇2+β22)E = 0 ( ∇2+β22)B = 0 // region 2 ( ∇2+β32)E = 0 ( ∇2+β32)B = 0 // region 3 (1.5.27) These last equations follow from the previous set using Ji = σiE and curl Ji = σi curl E = -jωσ iB. The solution method was outlined earlier: for each inhomogeneous equation compute the particular solution as a Helmholtz integral, then for all equa tions identify generic homogeneous solutions with unknown constants, and finally dete rmine those constants using bounda ry conditions from box (1.1.50). The potential approach has the advantage of a singl e wave operator and only two equations, while the damped field approach has the advantage of not in volving any currents, but the disadvantage of having three times more equations and requiring computation of grad ρ i. There is a lot more to keep track of. These are all of course vector Helmholtz equations. In a problem having only a single region (having μ,ε,σ) containing current density J and charge density ρ (perhaps inside the region, perhaps just on the surface), the Lorenz-gauge poten tial wave equations above in (1.5.25) may be written ( ∇ 2+k2)φ = -(1/ε) ρ k2 = ω2με all of region R ( ∇2+k2)A = - μJ . k2 = ω2με all of region R (1.5.28) Chapter 1: Basic Equations 58 These equations may be derived directly from the single-region field wave equati ons (1.2.1) and (1.2.2) converted to the frequency domain, ( ∇2 + k2)E = jωμ J + (1/ε) grad ρ k2 = ω2με ( ∇2 + k2)B = - μ curl J . (1.5.29) Each of these last four equations has its own Helmholtz integral, φ(x,ω) = 1 4πε ∫ρ(x',ω)e-jkR R dV' R = | x - x'| k2 = ω2με A(x,ω) = μ 4π ∫J(x',ω) e-jkR R d V ' ( 1 . 5 . 3 0 ) E(x,ω) = - 1 4π ∫[ jωμJ(x',ω) + (1/ε) grad ρ (x',ω) ] e-jkR R d V ' B(x,ω) = μ 4π ∫[ curl J (x',ω)] e-jkR R d V ' ( 1 . 5 . 3 1 ) The A(x,ω) Helmholtz integral (1.5.30) appears on Jackson p 408, Eq. (9.3), with j → -i and μ→ μ0. For this same single-region problem, the damped wave equation (1.5.27) becomes ( ∇ 2+β2)E = (1/ε ) grad ρ β2 = ω2μξ ( ∇2+β2)B = 0 ( 1 . 5 . 3 2 ) where again ρ might be in the volume and/or on the surface of the volume. This follows directly from (1.5.29) using the methods above. (e) Self Consistency of Helmholtz Integral Solutions The various Helm holtz partial differential equations encountered in the previous sections have solutions expressed as "Helmholtz integrals". In particular, ou r King gauge Helmholtz integrals for the potentials have this form, φ(x,ω) = 1 4πξd Σi∫nci(x',ω) e-jβdR R dS' R = | x - x'| (1.5.13) A (x,ω) = 1 4π Σi∫μiJi(x',ω) e-jβdR R d V ' . (1.5.9) These equations sometimes give the impression that one can willy-nilly specify an arbitrary charge distribution n ci and an arbitrary current distribution Ji for a set of transmission line conductors and then these Helmholtz integrals will generate the correct potentials A and φ from which the correct fields E and B may be obtained using (1.3.1), Chapter 1: Basic Equations 59 B = curl A E = - grad φ - jωA . (1.3.1) This is a false impression for one to infer from the discussion of the previous sections. For example, in a "fat twinlead" transmission line of the kind to be mentioned in Section 2.5 below, Fat twinlead Fig 2.16 the charge and current densities are extremely non- uniform. One cannot arbitrarily specify for this problem a uniform n and J z distribution in each conductor and expect the resultant E and B fields to be correct. The issue here is that solutions ha ve to be self-consistent. Suppose one were to specify for the above fat twin-lead problem a uniform n and J z. That is to say, one specifies that surface charge n is uniform around each circular cross section perimeter, and J z is uniform across each disk area. The Helmholtz integrals shown above would then yield some A and φ and that in turn would yield some E and B for the fields in the dielectric between the c onductors. One could then compute from the E field the value of surface charge n on each conductor using (1.1.47) n = εdEn, where E n is the normal E field just above the conductor surface. Similarly, one could compute c onduction currents in the conductors perhaps from J = (1/μ)curl B - jωε E which is Maxwell (1.1.1). One would find, unfortunately, that the resulting n and J did not agree with the initially assumed values of n and J. Such a "solution" is then meaningless because it is not self-consistent. All real-world Maxwell equation problems tend to have this circular aspect which makes solutions more difficult than the solution of idealized problem s. A problem mentioned elsewhere in this document is that of a radiating dipole antenna. One can assume a certain sine shaped current pattern in the antenna, compute from it the potentials and fields, and one w ill find when the antenna current is back-computed from those fields that the pattern is not quite a sine pattern unless the wire is infinitely thin. There are then two useful conclusions to be drawn here. First, if transmission line conductors are very thin relative to their spaci ng, it is just fine to assume a uniform charge and current distribution in those wi res, since the actual non-uniformity will have only a small effect on the solutions. Second, a general method of solution is to start with some charge and current distributions that seem reasonable based on one's general analysis of a problem. One can then find th e back-computed charges and currents, and adjust the input model accordingly. This would be the basis of either an analytic iterative procedure, where the model has some adjustab le parameters, or of a numerical procedure where the model is the set of values that comprise the charge and current distribution and some kind of iterative "relaxation" method then produces self-consistent solutions. We note that the exact solution of the "fat twin lead" transmission line is derived in Chapter 6 by a method which bypasses this iterative process, and whic h works only due to the simple nature of the geometry. Chapter 1: Basic Equations 60 1.6 Reinterpretation of all equations in terms of complex functions It see med useful to defer the topics of this section to avoid cluttering up the preceding five sections. The Fourier Transform has already been used in the pr evious two sections, and here we discuss it more formally as a motivating factor in changing our point of view from real to comp lex functions. The general nature of the Fourier Transform of complex monochrome (ejωt) fields sets the stage for the analysis of the round wire in Section 2. (a) Complex Functions Up to this point, we have been regarding the follo wing fields as representing real physical quantities, H(x,t) D(x,t) J(x,t) A(x,t) B(x,t) E(x,t) ρ(x,t) φ(x,t) . (1.6.1) The fields, potentials and sources exist in the real physical world and are related by equations involving real operators like curl and ∂/∂t. We can represent such an equation as L x,tf(x,t) = g(x ,t) where L x,t is some real differential operator and f and g are real fields. One can extend f and g such that f and g are eith er both the real or both the imaginary parts of complex functions F and G. Then the equation L x,tF(x,t) = G( x,t) represents two distinct physical equations which we can write as L x,tF(x,t) = G( x,t) => L x,t[f(x,t) + jf '(x,t)] = [g( x,t) + jg'( x,t)] => L x,t f(x,t) = g(x,t) F( x,t) = f(x ,t) + jf '(x,t) Lx,t f '(x,t) = g'(x,t) G( x,t) = g( x,t) + jg'( x,t) . (1.6.2) It is convenient to regard all the mathematical fields listed above in (1.6.1) as complex fields like F and G. For example, we might write the Maxwell curl E equation (1.1.2) in this manner curl E(x,t) = - ∂B(x,t)/∂ t E(x,t) = e(x,t) + j e'(x,t) B(x,t) = b(x,t) + j b'(x,t) . (1.6.3) where e = Re(E) and e' = Im( E) and similarly for the B field. The single left equation of (1.6.3) then represents these two different physical equations with real fields curl e(x,t) = - ∂b(x,t)/∂ t curl e'(x,t) = - ∂b'(x,t)/∂ t . ( 1 . 6 . 4 ) Thus, one can regard one's physical fields as either the real or imaginary parts of the complex fields. Chapter 1: Basic Equations 61 (b) Monochrome time The classic application of this idea is the assumption that some complex field is "monochrome" (monochromatic) in its time dependence, meaning for example, E i(x,t) = ej[ω1t+φi(x,ω1)] Ei(x,ω1) = ejω1t ejφi(x,ω1) Ei(x,ω1) , (1.6.5) where E i(x,ω1) = | Ei(x,t) | is real. Index i denotes a field component in an arbitrary coordinate system, not just Cartesian coordinates. All time dependence is in the ejω1t factor and all spatial dependence is in the factor [ejφi(x,ω1) Ei(x,ω1)] -- separation of variables. This monochrome field might be regarded as a probe or driver of some system and the solution fields E i(x,t) and phases φ i(x,ω1) might depend parametrically on the probe frequency ω1 as well as on position x. For (1.6.5) the corresponding physical field assumption is either of these equations, ei(x,t) = Re{ Ei(x,t)} = cos[ω 1t + φi(x,ω1)] Ei(x,ω1) e'i(x,t) = Im{ Ei(x,t)} = sin[ ω1t + φi(x,ω1)] Ei(x,ω1) . (1.6.6) We stress again that the phase φ i(x,ω1) might depend on both x and ω1. A good prototype 1D example for the ω1 dependence of phase φ1(x,ω1) is a damped harmonic oscilla tor with resonant frequency ω0 which is driven at frequency ω1. The solution is: x(t) = x(0) sin[ ω1t + φ(ω1)] tan φ(ω1) = -(ω1/τ)/(ω02- ω12) . Of course the solution function x(t) is not a field over R3, so in this case the phase φ has no x dependence. Comments: 1. The assumed form (1.6.5) is the most general form one can have for a monochrome field. One can always assume a more restrictive form for a certai n type of problem and see where it leads. Such a restricted form is an "ansatz" form meaning that one assumes that restricted form and then one tries to find the solution to a specific problem with the E field so restricted. If a solution is found which satisfies Maxwell's equations, then the ansatz form is justified. For example, one might use the more restrictive ansatz where φ i(x,ω) = φi(ω), or even more restrictive with φi(x,ω) = φi, a constant. 2. For a wave problem, one might try the following ansatz form which is a restriction of (1.6.5), E i(x,y,z,t) = ej(ω1t-kz) ejφi(x,y,ω1)] Ei(x,y,ω1) (1.6.7) where Ei(x,y,ω1) is real. In this form the entire dependence on t and z is exposed in the first factor, so the solution then represents a wave traveling in the z direction. Chapter 1: Basic Equations 62 3. Note in (1.6.5) that the phase function φi(x,ω1) can be different for different components E i(x,t). Appendix D studies the fields inside a round wire and the three field components E z, Er and Eθ do indeed have different phases for that problem. (c) Why complex fields: The Fourier Transform The reason for using a com plex field like E(x,t) instead of the real field e(x,t) has to do with the Fourier Transform (or the Laplace Transform). This transfor m is almost always need ed to solve a non-trivial problem involving Maxwell's equations, and we saw it in action in Section 1.5. With the convention that the (1/2π) goes in the expansion formula along with e+jωt, we write the Fourier Integral Transform as : [ for want of a better notation, f^(ω ) is the transform of f(t) ] E^(x,ω) = ∫-∞ ∞ dt E(x,t) e-jωt projection = transform (1.6.8a) E(x,t) = (1/2π) ∫-∞ ∞ dω E^(x,ω) e+jωt . expansion = inverse transform = recovery (1.6.8b) Here E(x,t) is the original complex field whose real and im aginary parts are physical fields as in (1.6.3) or (1.6.6), while E^ (x,ω) is the Fourier Transform of E( x,t). As (1.6.8) shows, the dimensional units of the F ourier transform of some quantity have an extra sec factor. For example, since dim[E(x ,t)] = volt/m, it follows that dim[ E^(x,ω)] = volt-sec/m. An obvious property of the Fourier Transform is this: ∂tE(x,t) = (1/2 π) ∫-∞ ∞ dω E^(x,ω) ∂t e+jωt = (1/2π) ∫-∞ ∞ dω [jω E^(x,ω)] e+jωt which we can write as ( symbol ↔ means "corresponds to") E(x,t) ↔ E^(x,ω) ⇔ ∂tE(x,t) ↔ jω E^(x,ω) (1.6.9) which is just another way to state our rule (1.5.2). In the case of assumed monochrome time dependence of the form (1.6.5) ( refl ected in (1.6.6) ) one finds that E i(x,t) = ej[ω1t+φi(x,ω1)] Ei(x,ω1) (1.6.5) E^i(x,ω) = ∫-∞ ∞ dt [ejω1t ejφi(x,ω1)Ei(x,ω1)] e-jωt = Ei(x,ω1) ejφi(x,ω1) ∫-∞ ∞ dt ej(ω1-ω)t = [ Ei(x,ω1) ejφi(x,ω1)] 2πδ(ω-ω1) ( 1 . 6 . 1 0 ) or E^(x,ω) = E( x,0) 2πδ(ω-ω1) . (1.6.11) It is this very simple single- δ-function form that motivates the use of complex fields as carriers of the real physical fields. One can of course Fourier-transfo rm the monochrome physical field directly, but the result is clumsy to deal with. For example, Chapter 1: Basic Equations 63 ei(x,t) = Re{ Ei(x,t)} = cos[ω 1t + φi(x,ω1)] Ei(x,ω1) e'i(x,t) = Im{ Ei(x,t)} = sin[ ω1t + φi(x,ω1)] Ei(x,ω1) . (1.6.6) e^i(x,ω) = ∫-∞ ∞ dt { cos[ω1t + φi(x,ω1)] Ei(x,ω1) }e-jωt = Ei(x,ω1) (1/2) ∫-∞ ∞ dt { ej[ω1t+φi(x,ω1)] + e-j[ω1t+φi(x,ω1)] } e-jωt = Ei(x,ω1) [ejφi(x,ω1)πδ(ω-ω1) + e-jφi(x,ω1)πδ(ω+ω1) ] . (1.6.12) or e^(x,ω) = [e(x,0) + j e'(x,0)] π δ(ω-ω1) + [ e(x,0) - j e'(x,0)] π δ(ω+ω1) . (1.6.13) This lacks the friendliness of (1.6.11) in that the real and imaginary parts of E(x,0) both appear on the right, and two different ω-space delta functions are required. One could by fiat set e' = 0, for example, but the two delta functions still remain. A directly related benefit of using the complex functi on approach is the fact that math with exponentials is so much simpler than the corresponding math with trig functions, as for example e j(ωt+φ) e-j(ω't+φ') = ej(ω-ω')t ej(φ-φ') // dependence on t isolated to one factor versus cos(ωt+φ)cos(ω't+φ') = (1/2) { cos[ ( ω-ω')t + (φ-φ')] + cos[ ( ω+ω')t + (φ+φ')] } . Comment: Using the real cosine form shown as the first line of (1.6.6) along w ith the Fourier Cosine Transform is not viable because cos[ ω1t + φi(x,ω1)] Ei(x,ω1) is not an even function of t. (d) Monochrome E and B fields One might seek to solve a system using monochrome fi elds of the form (1.6.5) for both the electric and magnetic fields. Those forms would be ( E and B are real) Ei(x,t) = ej[ω1t+φei(x,ω1)] Ei(x,ω1) Bi(x,t) = ej[ω1t+φbi(x,ω1)] Bi(x,ω1) ( 1 . 6 . 1 4 ) where we assume the same frequency ω1 for both fields, but allow the fields to have different phase functions φ ei and φbi. In this case (1.6.10) becomes E^i(x,ω) = Ei(x,ω1) ejφei(x,ω1) 2πδ(ω-ω1) B^i(x,ω) = Bi(x,ω1) ejφbi(x,ω1) 2πδ(ω-ω1) . (1.6.15) The ratio of E i over Bj is then given by E^i(x,ω) B^j(x,ω) = Ei(x,ω1) Bj(x,ω1) ej[φei(x,ω1)- φbj(x,ω1)] . (1.6.16) Chapter 1: Basic Equations 64 Since Ei and Bj are real, the phase of the ratio E^ i/ B^j is determined by the last factor and will in general be a function of both position x and frequency ω1. We shall see this situation arise in Chapter 2 when we calculate the fields inside a conducting round wire . (e) A Pitfall to Avoid Notice that E(x,t) = e(x,t) + j e'(x,t) => E^(x,ω) = ∫-∞ ∞ dt E(x,t) e-jωt = ∫-∞ ∞ dt [e(x,t) + j e '(x,t)] e-jωt = e^(x,ω) + j e'^(x,ω) . ( 1 . 6 . 1 7 ) Whereas e(x,t) and e'(x,t) are the real and imaginary parts of E(x,t), the functions e^(x,ω) and e'^(x,ω) are not the real and imaginary parts of E^(x,ω) since in general e^(x,ω) and e'^(x,ω) are both complex functions. In this document we shall neve r deal with transforms of the type e^(x,ω) or e'^ (x,ω). (f) Overloaded Notation and Maxwell's Equations in ω space In this section we have carefully deno ted the Fourier Transform of f(t) as f^( ω) which is a notation used by Stakgold and others (though Stakgold has our (1.6.8) phases negated as in his equation (5.32) ). In the rest of this document, however, we represent the Fourier Transform of f(t) as f(ω ) to avoid a proliferation of hat ^ symbols. Since the functions f(t) and f( ω) are completely different functions, the symbol f is "overloaded" (in the sense of overloaded variable nam es in computer languages) and we trust the reader to understand that f( ω) always means f^( ω). It is the presence of the argument ω that cues the reader to this fact. This overloaded notation has already been us ed in Section 1.5 and we continue it right here: In the Maxwell and related equations which include the ∂ t operator, if the fields are expanded onto their Fourier transformed components using (1.6.8b), then using the rule (1.6.9) one may instantly write the frequency-domain version of these equations, just as in the example of Section 1.5. For example, curl H(x,ω) = jωD(x,ω) + J (x,ω) ( 1 . 6 . 1 8 ) curl E(x,ω) = -jωB(x,ω) ( 1 . 6 . 1 9 ) div J(x,ω) = -jωρ(x,ω) . ( 1 . 6 . 2 0 ) Other equations in the Section 1.1 list have the same form but in terms of the frequency-domain functions. For example, J(x,ω) = σ (x) E(x,ω) ( 1 . 6 . 2 1 ) where we momentarily allow σ(x) to have spatial dependence but not time dependence. Chapter 2: The Round Wire and the Skin Effect 65 Chapter 2: The Round Wire and the Skin Effect Chapter 1 dealt with the generalities of electromagnetic theory. Maxwell's equations were stated ex machina , as it were, and wave equations fo r the fields and potentials were then derived. Formal integral solutions of the potential wave equations were also derived using the Green's Function method. It was noted that the potentials φ and A are parts of the same Lorentz 4-vector. Whereas the approach of Chapter 1 was very general and abstract, the discussion of this chapter is highly specific. The goal here is to learn about the prop erties of a very simple object -- an infinite straight round wire. Although transmission lines are not always made out of round wires, there is a wealth of useful practical information that arises from the stud y of this simple example which applies to more general geometries. The major issue here is called the "skin effect". At high frequencies, current is forced away from the central regions of a conductor and concentrates at the su rface in a thin layer that has a characteristic depth called δ, the skin depth. In this chapter it will be show n exactly why this occurs. The significance of the effect is that the resistance (impedance) of a wire increases drastically at high frequency since the current is forced to flow only in this thin shell below the wire surface. This effect is manifested in a property of a wire called its surface impedance which is studied belo w in Section 2.4 and qualitatively in Section 2.5. Our development is an extension of the excellent discussion of Matick's Chapte r 4. It is fastest to solve the round wire problem starting with the ω-domain damped wave equati on (1.5.32) which, inside the wire where there is no free charge, says (∇ 2 + β2)E = 0 with β2 = ω2μξ where μ and ξ apply to the conductor . Instead, we have chosen to start from the basic Maxwell curl equations and use simple "math loops" to derive the basic (first or der differential) equations relating E and B fields. The general technique of putting loops in op portune places is extremely useful in an alyzing the more complicated situation which arises in a transmission line. This method is ca rried out in Section 2.2 and the wire's interior solutions are then studied in Section 2.3. In the work done below, we shall assume axia l symmetry for the fields in the round wire. The problem is treated more generally in Appendix D where the Helmholtz equation ( ∇2 + β2)E = 0 is directly solved. The partial wave m = 0 solution of A ppendix D corresponds to the analysis below. 2.1 The Implicit Wave Context, Helmholtz Equations and the Skin Effect In the sections below we don' t explicitly consider the notion that a wave is trav eling down our round wire, but that is in fact what is happening and this f act deserves a few comments before we delve into the interior solution of the wire. Specifically, imagine that E has the following traveling-wave form, E(x,y,z,t) = ej(ωt-kz) E(x,y,ω) ( 2 . 1 . 1 ) where E(x,y,ω) might have dependence on ω and might be complex. Inside the wire this field must satisfy the damped wave equation (1.3.36), ( ∇ 2 - με ∂t2 - μσ∂t) E(x,y ,z,t) = 0 . (2.1.2) When the form (2.1.1) is inserted into (2.1.2), the result is, using ∇2 = ∇2D2 + ∂z2, Chapter 2: The Round Wire and the Skin Effect 66 ( ∇2D2 + ∂z2 - με ∂t2 - μσ∂t) { ej(ωt-kz) E(x,y,ω)} = 0 or ( ∇2D2 - k2 + με ω2 - jμσω ) { ej(ωt-kz) E(x,y,ω)} = 0 or ( ∇ 2D2 - k2 + β2) { ej(ωt-kz) E(x,y,ω)} = 0 or ( ∇ 2D2 + β2 - k2) E(x,y,ω) = 0 // a 2D Helmholtz equation (2.1.3) where β2 = μεω2 - jωμσ = ω2μ (ε - jσ/ω) = ω2μ ξ . ξ ≡ ε - jσ/ω (1.5.1c) Comment : E(x,y,ω) is proportional to the Fourier Transformed ω-domain version of E(x,y,z,t) : E^ (x,y,z,ω ') ≡ FT{ E(x,y,z,t), ω'} = e-jkz E(x,y,ω) 2πδ(ω-ω') . // see (1.6.11) A similar equation applies just outside the round wire in the dielectric medium in which it is embedded, and this medium has its own β which we call β d. Thus we have ( ∇2 2D + β2 - k2) E(x,y,ω) = 0 inside wire β = (j - 1) ωμσ/2 = complex (2.1.4a) ( ∇2 2D + βd2 - k2) E(x,y,ω) = 0 outside wire βd = ω μdεd = ω/vd ≈ real . (2.1.4b) We have assumed that • the dielectric is non-conducting or only slightly conducting so ξd ≈ εd. • the conductor is a good one, so β2 ≈ (- j) ωμσ and then β = -j ωμσ . As will be shown below, the choice for -j is ej3π/4 = (j-1)/ 2 which then gives β = (j - 1) ωμσ/2 as in (2.1.4a). In (2.1.4b) we then make the ansatz assumption that k = βd ( 2 . 1 . 5 ) which basically says that the wave form e j(ωt-kz) E(x,y,ω) really does describe a wave traveling down the wire with k = βd. This k = βd = ω/vd is then related to the speed of light in the dielectric and is the expected value of k for, say, a radio or light wave traveling through the dielectric with no wire present. Once we have assumed ej(ωt-βdz) for the dielectric solution, the boundary conditions (1.1.50) on field components at the wire surface will force this sa me dependence on the solution inside the wire. In (2.1.4a), since the conductor has such a large σ, |β| is a large number and | β| >> βd (unless ω is very large or, as we shall later see, very small), so k2 can be ignored in (2.1.4a). We then have ( ∇2 2D + β2) E(x,y,ω) ≈ 0 inside wire β = (j - 1) ωμσ/2 = complex (2.1.6a) ∇2 2D E(x,y,ω) = 0 outside wire (2.1.6b) Chapter 2: The Round Wire and the Skin Effect 67 The second equation says that the E field outside the wire must solve the 2D vector Laplace equation. See Appendix D.7 for comments on the general exterior solution. Appendix D uses β '2 ≡ β2 - βd2 and does not make the approximation that β' ≈ β, but we make that approximation here. To put our "isolated" round wire into a physical context, it helps to think of it as the round central conductor of a coaxial cable whose shield cylinder radius is very large compared to the central wire radius (the Great Cylinder, analogous to the Great Sphere of el ectrostatics). Then this central round wire is really part of a transmission line and we expect such a transmission line to carry a wave with ej(ωt-βdz) time and z dependence. Moreover, we expect the field soluti on inside such a coaxial cable central wire to have the axial symmetry that appears in our assumption list below. It is equation (2.1.6a) for the wire interior that we shall encounter below, and hopefully we have now put that equation into the context of a wave traveling down the wire. If β d has a small negative imaginary part due to conductiv ity of the dielectric (see (1.5.1a)), the factor e-jβdz says that the wave slowly damps out as it trav els down the wire due to dielectric ohmic loss, as it well should. (In a laser inverted medium βd has a positive imaginary part so the wave grows instead.) On the other hand, β is huge and has equal real and imaginary part magnitudes. Due to our axial symmetry, (2.1.6a) really says ( ∇2 2D + β2) E(r,ω) = 0 which can be thought of as a "wave equation" in the radial direction. Of course it is a damped wave equation of a very extreme sort. As one moves in from the surface of the wire toward the center, we show later that over a distance in which the "wave" phase changes by about π /2, the amplitude is already down by a factor 1/e, so one can roughly say that the wave basically damps out before it even goes 1/2 wavelengt h. This is the skin effect described below. To understand this effect, it is useful to consider a 1D version of the situation. Imagine zooming the camera in very close to the left surface of the round wire's cross section, so that we see a half space of conductor on the right and a half space of dielectric on th e left. Let the radial direction be called x which increases into the conductor with x = 0 at the interf ace. Then the inside-wire wave equation above says (∂ x2 + β2) E(x,ω ) = 0 . ( 2 . 1 . 7 ) The solution to this equation is (we select a pa rticular sign for the phase, and see (2.1.6a) for β) E(x,ω) = E(0,ω ) e+jβx = E(0,ω)exp{ j [(j - 1) ωμσ/2 ] x} = E ( 0 , ω) exp{- ωμσ/2 x} exp{ -j ωμσ/2 x} = E ( 0 , ω) exp{- x/ δ} exp{ -j x/ δ} δ ≡ 2/(ωμσ) = E ( 0 , ω) e-x/δ e-jx/δ . // = E(0, ω) e-(j+1)x/ δ (2.1.8) As one moves from x=0 to the right into the conductor, in distance δ the E field amplitude drops to 1/e and the phase has changed by π/2. Quantity δ is called the skin depth, and this is probably the most basic way to understand the notion of the skin effect. It is a result dictated by the Helmholtz equation having a complex parameter β of the type shown. Based on this argument , the skin effect occurs at any conductor surface regardless of its cross-sectional shape. Below we see in the round wire example how the Helmholtz equation (2.1.6a) is in turn a result of th e two Maxwell curl equations each of which relates E and B. Of course this is how the Helmholtz equation w as derived in the first place starting with (1.2.1). Chapter 2: The Round Wire and the Skin Effect 68 2.2 Derivation of E(r), B(r) and J(r) for a round wire We convert (1.1.3 8) and (1.1.36) to the ω domain using rule (1.5.2), curl B = μ (jωε E + J ) ⇔ ∫{C B • ds = μ∫S [jωεE + J ] • dS (2.2.1) curl E = − jωB ⇔ ∫{C E • ds = -jω∫S B • dS . (2.2.2) The two terms on the right side of (2.2.1) have names (we sometimes omit the word "density") jωεE = displacement current (density) // amps/m 2 J = σE = conduction current (density) // amps/m2 . The sum of both currents may be written as ( jωε + σ) E(x,ω ) . For any metal conductor such as copper, the displacement term is completely negligible as long as ωε << σ. The value of ε for a metal is not very obvious and is likely in fact to be negative at frequencies below optical frequencies (free electron gas, plasma frequenc y, Drude model, etc), so we will follow Matick p 118 and blindly set ε = ε 0 for a crude comparison. The condition for negligible displacement current ωε << σ then becomes f << σ/[2πε0]. Using σ = 5.81 x 107 mho/m and ε0 = 8.85 x 10-12 F/m, one gets f << (σ /2πε ) = 1.04 x 1018 Hz ≈ one billion GHz Therefore, the displacement current is always ignored inside a conductor for any conventional transmission line application. Whatever ε really is, we shall ignore j ωε compared to σ. All the current inside a good conductor is conduction current. With r espect to (1.5.1c), this same approximation means that inside a conductor, ξ ≈ - jσ/ω β 2 ≈ - jωμσ // f << 1018 H (2.2.3) We now make a set of assumptions: (a) the round wire conductor medium is uni form (homogeneous) and isotropic (2.2.4) (b) the current pattern in the wire is axially symmetric (no dependence on azimuth θ; it is invariant under any rotation of the wire about its center line) (c) the current is axial (longitudinal), so J = Jz^ , so J = J z (d) the E field is also axial so E = E z^ ( this follows from (c) and J = σE ), so E = E z (e) The B field lines go around in circles centered at the wire axis. The relation between the direction o f B and the current flow J is given by the right hand rule. If J = J z > 0, then B = B θ > 0. Thus, we represent J z(x,y,ω) = J(r), E z(x,y,ω) = E(r), and B θ(x,y,ω ) = B(r) -- no dependence on θ or z. The fields like E z(x,y,ω) are of the type shown on the right of (2.1.3) where the z dependence has already Chapter 2: The Round Wire and the Skin Effect 69 been extracted. Fields E(x,y, ω), B(x,y,ω ) and J(x,y,ω) are complex, so E(r), B(r) and J(r) are all complex. They all depend on ω, but we suppress the ω arguments. As noted above, E = E(r) z^ and B = B(r) θ^. Here is another way to state assumption (b). We search for an axially symmetric solution of Maxwell's equations for the round wire, and if we find one, we accept it as a possible way fields and currents could exist in the wire. If the wire were in idealized perfect isolation with an axially symmetric source and load, the invariance of the physical situati on with regard to rotation about the wire axis would require (b) to be valid. This symmetry is also imp lied by our "fat" coaxial cable context noted earlier. Consider now the thin (width is dr) red loop shown in Fig 2.1: Fig 2.1 Cross section view of wire, current flowing in z^ direction toward viewer According to (2.2.1) with J = σE and no displacement current, ∫{C B • ds = (μσ ) ∫S E • dS . ( 2 . 2 . 5 ) For the CCW loop shown, the "right hand rule" says area dS points out of the plane of paper. The two sides of this equation can be easily evaluated (B = B θ and E = E z) [ B(r+dr) (r+dr) - B(r) r] θ = (μσ) E(r) [ rθ dr ] (2.2.6) which simplifies to ∂ [r B(r)] ∂r = (μσ) [r E(r) ] . (2.2.7) Comment : When one says in Fig 2.1 that " J points in the z^ direction", one interpretation might be that the vector J has the form J = Jzz^ and that Jz > 0. That is not the correct interpretation for our pictures. The quoted phrase just means that J = Jzz^ and nothing is implied about the "sign" of J z. In our case, J z = J(r) is a complex number which has no "sign". If we said " J points in the - z^ direction" we would just mean that J = Jz(-z^) = - Jzz^. Our only interest in clarifying these "directions" is to get the signs right in our application of Stokes's law. The same comment applies to the direction of B in the next figure. By saying that " B points out of the plane of paper", we just mean that B = +B(r) θ^ which is consistent with the fact that J = +Jzz^ according to the right hand rule: thumb in the "direction" of J at the wire axis, curled fingers are in the "direction" of B. Chapter 2: The Round Wire and the Skin Effect 70 Now consider the thin red loop shown in Fig. 2.2, Fig 2.2 Top view of wire's central plane, current flowing in z^ direction (down) According to (2.2.2), ∫{C E • ds = -jω∫S B • dS . (2.2.2) where, for the CCW loop shown, the right hand rule puts dS pointing to the viewer (aligned with B which points to the viewer due to its right hand rule with J ). The two sides of this equation are easily evaluated (the first term on the left is negative because z^ points down while the red arrow points up) [ - E(r+dr) + E(r)] s = -j ωB(r) [ s dr ] (2.2.8) which simplifies to ∂E(r) ∂r = jωB(r) . (2.2.9) This equation says that E(r) changes with radius as long as ω ≠ 0 and B(r) ≠ 0. Since everything is complex, we cannot really tell from (2.2.9) that |E(r)| increases with radius, but we shall see below that it does, and this fact gives rise to the "skin effect " where current is maximum at the wire surface. Reader Exercise : Why can't one take the absolute value of both sides of (2.2.9) and reach the conclusion that ∂ r|E(r)| = ω|B(r)| > 0 and conclude that |E(r)| increases with r? Hint: ∂x|f| ≠ |∂xf| Now solve (2.2.9) for B(r) and put this into (2.2.7) to get 1 r ∂ ∂r (r ∂E(r) ∂r ) = (jωμσ) E(r) = - β2 E(r) (2.2.10) where β2 = - jωμσ from (2.2.3). The operator on the left is ∇2 in cylindrical coordinates for a function that does not depend on θ or z. For such functions, ∇2 = (∂z2 + ∇2 2D) and ∇2 2D are equivalent. Thus, (2.2.10) is really a special case of the following Chapter 2: The Round Wire and the Skin Effect 71 [ ∇2 2D + β2 ] E(r) = 0 . (2.2.11) This in turn is a special case of the E field wave equation (2.1.6a), ( ∇2 2D + β2) E(x,y,ω) = 0 . (2.1.6a) The Helmholtz parameter β2 is given by (1.5.1c) , β2 = μεω2 - jωμσ = ω2μ ( ε - jσ/ω) ≈ -jωμσ . (1.5.1c) The ε term in β2 has been neglected since σ is very large. Parameter β = 2π/λ is a "wavenumber" and has dimensions of m-1. If λ were real (it is not), then β would be the number of wave radians per meter just as ω is the number of wave radians per second. The complex number -j has two square roots which are e j3π/4 and e-jπ/4, Fig 2.3 and we specify the upper red arrow as th e square root in the definition of β , β ≡ ej3π/4 ωμσ . // β = (j - 1) ωμσ/2 = (j-1)/ δ (2.2.12) We could have started out with (2.2.11) and skipped all the above analysis of loops, but this method of using loops emphasizes the direct action of Ma xwell's equations and seems instructive. Comment : One could of course take the other square root of -j and develop things that way. Historically the root selected above has been used. Taking the other root means β → -β. A review of the solutions obtained below shows that they are invariant under β → -β. Such a review can use the facts that J 0(-z) = J0(z), and J 1(-z) = -J 1(z). In general J ν(z) is analytic at z = 0 for Re ν ≥ 0 and the rules just stated follow from the series representations of J 0 and J1 as shown for example in Spiegel 24.5 and 25.6. In "exterior" problems involving the Hankel functions, there is significance as to whether z = βr is in the upper or lower z-plane in terms of convergen ce for large r. For example, if β is in the upper half plane, then H(1)(βr) is the function that converges as r→∞ and H(2)(βr) blows up: Chapter 2: The Round Wire and the Skin Effect 72 http://en.wikipedia.org/wiki/Bessel_function The next step is to expand (2.2.10) as follows: ∂ 2E(r) ∂r2 + 1 r ∂E(r) ∂r + β2 E(r) = 0 or r2 ∂2E(r) ∂r2 + r ∂E ∂r + r2β2 E(r) = 0 . (2.2.13) Change variables to dimensionless x = βr. Then r = x/β x = βr ∂x/∂r = ∂rx = β ∂rE = ∂E ∂r = ∂E ∂x ∂x ∂r = β ∂E ∂x ( 2 . 2 . 1 4 ) ∂2E ∂r2 = ∂r2E = ∂r(∂rE) = ∂r(β∂xE) = β∂x(∂rE) = β∂x(β∂xE) = β2∂x2E = β2 ∂2E ∂x2 . (2.2.15) Inserting these quantities into (2.2.13) gives r 2β2 ∂2E ∂x2 + r β ∂E ∂x + r2β2E = 0 or x 2 ∂2f(x) ∂x2 + x ∂f(x) ∂x + x2f(x) = 0 (2.2.16) where f(x) = E(x/ β). Now (2.2.16) happens to be Bessel's Equation with ν = 0 [NIST 10.2.1], and the solution must therefore be a linear combination of this form, where C and D are constants, f(x) = C J 0(x) + DY 0( x ) . ( 2 . 2 . 1 7 ) So far, we still don't know which way ∂E/∂r in (2.2.9) is changing, but we are about to find out. Since f(x) represents the current and the electric field, we know f(0) cannot be infinite. But Y 0(x) blows up at x=0 [NIST 10.8.2] , therefore constant D = 0. We now have an exact solution for the electric field in the wire: E(r) = f(x) = C J 0(βr ) ( 2 . 2 . 1 8 ) where β = ej3π/4 ωμσ = (j - 1) ωμσ/ 2 ( 2 . 2 . 1 9 ) Chapter 2: The Round Wire and the Skin Effect 73 The following definition is usually made (factor of 2 explained later) δ ≡ 2/ωμσ = skin depth // ωμσ = 2/δ2 (2.2.20) so that β = e j3π/4 (2 /δ) = (j-1)/δ and β2 = -2j/δ2 . (2.2.21) It is convenient to divide (2.2.18) by itself evaluated at r=a which we shall assume is the radius of our round wire, so (plots coming soon), E(r) = E(a) J0(βr) J0(βa) . ( 2 . 2 . 2 2 ) According to (2.2.9) which says ∂rE(r) = jωB(r) we can write B(r) = (1/j ω)∂rE(r) = (1/j ω) E(a) ∂r[J0(βr)] J0(βa) = (1/jω) E(a) βJ0'(βr) J0(βa) = (β/jω) E(a) J0'(βr) J0(βa) . Since J0'(x) = -J1(x) [ NIST 10.6.2 ] this gives, B(r) = -( β/jω) E(a) J1(βr) J0(βa) ( 2 . 2 . 2 3 ) which when evaluated at r = a gives B(a) = -( β/jω) E(a) J 1(βa) J0(βa) => E(a) = - (j ω/β) J0(βa) J1(βa) B(a) . (2.2.24) which relates the two surface values B(a) and E( a). An alternative way to write B(r) is B(r) = B(a) J1(βr) J1(βa) = {-(β/jω) E(a) J1(βa) J0(βa) } J1(βr) J1(βa) = -(β/jω) E(a) J1(βr) J0(βa) . (2.2.25) Ampere's law with a circular loop just below the surface gives, where I is the total wire current, 2πaHθ = I => 2 πaBθ/μ = I => B(a) = μI 2πa (2.2.26) from which we find from (2.2.24) that E(a) = - (j ω/β) μI 2πa J0(βa) J1(βa) . (2.2.27) Chapter 2: The Round Wire and the Skin Effect 74 Using (2.2.26) for B(a) in (2.2.25) gives B(r) = μI 2πa J1(βr) J1(βa) ( 2 . 2 . 2 8 ) and using (2.2.24) for E(a) in (2.2.22) gives E(r) = E(a) J 0(βr) J0(βa) = {- (jω/β) μI 2πa J0(βa) J1(βa) } J0(βr) J0(βa) = - (jω/β) μI 2πa J0(βr) J1(βa) . (2.2.29) Let us gather up all the main results obtained so far and put them in a box: Interior Field Solution of a Round Wire (2.2.30) ∂ [ rB(r)] ∂r = (μσ) [ r E(r) ] (2.2.7) ∂E(r) ∂r = jω B(r) (2.2.9) B(r) = μI 2πa J1(βr) J1(βa) (2.2.28) B(a) = μI 2πa (2.2.26) E(r) = μI 2πa J0(βr) J1(βa) (-jω/β) (2.2.29) E(a) = μI 2πa J0(βa) J1(βa) (-jω/β) (2.2.27) J(r) = μI 2πa J0(βr) J1(βa) (-jωσ/β) = I 2πa J0(βr) J1(βa) β from J(r) = σE(r) β = ej3π/4 (2 /δ) = ej3π/4 ωμσ = (j - 1) ωμσ/2 = (j - 1)/ δ (2.2.19), (2.2.21) δ ≡ 2/ωμσ = skin depth β2 = -2j/δ2 = -jωμσ (2.2.20) , (2.2.21) The reader is reminded once again that E(r), B(r) and J(r) are complex functions of r and ω since they are components of the Fourier Integral Transform of the time-domain fields and current density. Since β is complex, the various J ν(βr) are also complex. Thus, the nature of the solutions in the above box is not very obvious at this point. Comment: Appendix D does an exact calculation of the E and B fields inside a round wire using a partial wave analysis with index m. The solution for the prob lem considered here in Ch apter 2 corresponds to the partial wave m = 0 and is stated in summa ry box (D.6.1). It is shown that the E z and Bθ fields there match those obtained here, but in addition there is an extra field component E r which is very small in the ratio |βd/β|. The reason our calculation here failed to disc over this smaller field component was that we assumed E = E(r) z^ and B = B(r) θ^. An implication of E r ≠ 0 is that E r(r=a) ≠ 0 which, as shown in Chapter 2: The Round Wire and the Skin Effect 75 (D.2.24), implies the existence of a surface charge on the round wire. This then fits with our context model of the round wire as the central conductor of a fat coaxial transmission line as discussed in Section 2.1. 2.3 A study of the solution of a round wire (a) Kelvin Functions The reader may be aware of the so-called first-kind modified Bessel function defined by Iν(x) ≡ e-jπν/2 Jν(ejπ/2x) , where the J ν function argument has phase π/2. Unfortunately, our J ν(βr) functions have phase 3 π/4 so the Iν functions are not particularly useful. The real and imaginary parts of a Bessel function having an argument with phase (3/4) π have the following historic names (bessel real and bessel imag inary) called Kelvin functions [ NIST 10.61.1 ], J ν(ej3π/4z) = ber ν(z) + j bei ν(z) . (2.3.1) In our application ej3π/4z = β r = ej3π/4(2 /δ) r so that z = 2 (r/δ). Thus, the solution E(r) in (2.2.22) may be written as, E(r) = E(a) ber0[2(r/δ)] + j bei 0[2(r/δ )] ber0[2(a/δ)] + j bei 0[2(a/δ)] . z = 2 (r/δ) (2.3.2) The Kelvin functions are real when the arguments ar e real and positive. This is the case for almost all special functions (they are "real analytic"), though derived functions like H n(1)(z) are an exception. Similar functions ker ν and kei ν are associated with K ν(ej3π/4z ) where K ν is the second-kind modified Bessel function. Since J(r) = σ E(r), we could replace E with J on both sides of (2.3.2). This equation for J appears in Matick as p 101 (4-18). Note: Lord Kelvin (William Thomson) introduced the ber and bei notation for these functions while considering the same problem we are dealing with he re. The functions appear in the Appendix of his 34 page 1889 inaugural address used when he became preside nt of the Institute of Electrical Engineers (see Refs) : We can verify using Maple (which Kelvin would have enjoyed) that these are the ber 0 and bei0 functions: Chapter 2: The Round Wire and the Skin Effect 76 Some authors, not liking Kelvin's notation, use Ber, Bei, Ker, Kei for ber, bei, ker, kei. Perhaps the idea is that Be is more obviously Bessel and perhaps Ke is then for Kelvin. Since these Bessel forms occur a lot, there are standard functions fo r their magnitude and phase [ NIST 10.68.1 ] J ν(ej3π/4z) = M ν(z) ejθν(z) . (2.3.3) Of particular interest is the magnitude of E(r). Applying (2.3.3) to the E(r) in (2.2.22) gives |E(r)| = |E(a)| M 0[2 ( r/δ )] M0[2 ( a/δ )] . (2.3.4) (b) Plots of |E(r)/E(a)| for various δ values Finally we are in a position to make some plots to see how the electric field magnitude varies with radius in a round wire as a function of the skin depth parameter δ ≡ 2/ωμσ . As ω increases, δ decreases. Our aging Maple V knows about the Kelvin functi ons but not M, so here is the code and here are plots of |E(r)| / |E(a)| for a = 20 and δ = 1 to 10, The steepest curve is for δ = 1: Chapter 2: The Round Wire and the Skin Effect 77 Fig 2.4 The same plots apply to |J(r)| / |J(a)|. One sees cl early how the current and electric field magnitude drop off quickly moving in from the edge of the round wire (right edge of graph) toward the wire axis when δ is small relative to radius a. Asymptotic expansions for M n(z) and θn(z) for large z are given by NIST 10.68.16 and 10.68.18, Mν(z) ≈ exp(z/ 2 ) 2πz [ 1 - 4ν2-1 82 z + O(1/z2) ] θν(z) ≈ (z/ 2 ) + (π/2) [ ν - 1/4 ] + 4ν2-1 82 z + O(1/z2) . (2.3.5) For ν = 0 we find M0(z) ≈ exp(z/ 2 ) 2πz [ 1 + 1 82 z ] θ0(z) ≈ (z/ 2 ) - (π/8) – 1 82 z . (2.3.6) For z > 3 the correction term 1 82 z in M0(z) is less than .03 so we can ignore it for rough estimates. In this case one gets |E(r)| = |E(a)| M0[2 ( r/ δ )] M0[2 ( a/ δ )] = |E(a)| M0(z) M0(za) z = 2 (r/δ) za = 2 (a/δ) ≈ |E(a)| exp(z/ 2 ) exp(za2 ) 2πza 2πz = |E(a)| exp([z-z a]/2 ) za/z . But [z-za]/2 = (r/δ)-(a/δ) = (r-a)/δ and za/z = a/r . Thus we find that Chapter 2: The Round Wire and the Skin Effect 78 |E(r)| |E(a)| = a r e(r-a)/δ r/δ > 3/ 2 = 2.1 . (2.3.7) This is the famous skin depth result as it appears for a round wire. This ratio is 1 at the surface and then drops off exponentially with characteristic distance δ moving inside the wire. One sees now why the 2 was included in the definition of δ: there is then no 2 in equation (2.3.7). Comparing (2.3.7) to the one- dimensional skin depth formula (2.1.8) one sees an extra a/r factor arising from the cylindrical geometry. Equation (2.3.7) is valid down to within about 2 skin de pths of the center axis of the wire. In general, one can assume the field E(r) is zero for all practical purpo ses perhaps 5 skin depths in from the surface (if a > 5δ). Here are plots of |E(r)|/|E(a)| using the approximate formula (2.3.7) for the same ten δ values as our previous plots, Fig 2.5 Previous plot using a certain approximation discussed above and here are the two sets of plots superi mposed with some notations added: Chapter 2: The Round Wire and the Skin Effect 79 Fig 2.6 The wire radius is a = 20, and the curves are for δ = 1 to 10, with δ = 1 being the rightmost and steepest curve. The red (exact) and black (approximate) curves for δ = 1 agree down to r = 2 at least. The δ = 5 red/black pair of curves start to pull apart around r = 10 which is 2 skin depths from the center. The δ = 8 red/black pair of curves start to pull apart around r = 16. We thus verify the claim made above that each red/black pair of curves agree starting at r = a and moving in to about 2 skin depths from the center line (the pull-apart points are marked by dots). One can al so see that the electric field is roughly zero about 5 skin depths in from the surface (marked by x's). Here are some skin depth valu es in copper based on (2.2.20) δ = 2/(ωμσ) with σ = 5.81 x 107mho/m, and μ = μ0 = 4π x 10-7 H/m. Selecting a reference point of 1 GHz, we have, δ = 2/(2πfμ0σ) = 1/(πfμ0σ) = 1/f 1/(π 109μ0 σ) = 2.09 x 1/f(GHz) μ (2.3.8) Here then is a table of copper skin depths ( μ = microns), f δ f δ 100 GHz 0.21 μ 100 KHz 209 μ 10 GHz 0.66 μ 10 KHz 661 μ 1 GHz 2.09μ 1 KHz 0.21 cm 100MHz 6.61μ 100 Hz 0.66 cm 10 MHz 20.9 μ 10 Hz 2.09 cm 1 MHz 66.1 μ 1 Hz 6.61 cm (2.3.9) Chapter 2: The Round Wire and the Skin Effect 80 The radius of the center conductor of Belden 8281 coaxial cable is 15.5 mil = 394 μ, so the skin effect restriction occurs for f ≈ 1 MHz and above. At 1 GHz δ is about 1/200th the radius. As we get into the lower frequencies, the exponential decay no longer applies for Belden 8281. For very low frequencies, we can use the small z limit of J 0(z) to see how the distortion begins at low frequency, J0(z) = 1 - z2/4 z << 1 // Spiegel 24.5 z = 2 (r/δ) . Using the expression for E(r) and β2 in box (2.2.26) we find E(r) E(a) = 1 + j (r/δ)2/2 1 + j (a/δ)2/2 |E(r)| |E(a)| = 1 + (r/δ )4/4 1 +(a/δ)4/4 . (2.3.10) This shows the very early phase of the skin effect ha ppening at low frequencies. Eq. (2.3.10) would apply for example in Belden 8281 at 1 KHz and below where δ/a ≥ 5. There is a very slight dip in the E(r) and J(r) distribution at r=0 compared to r=a. For example, with a = 20 and δ = 100 one has z ≤ 2 (a/δ) = 2 (1/5) = .28 for all values of r, so z is "small" in the whole range. Below is a plot of |E(r)/E(a)| in this case. Notice the offset zero so the drop is only 2 parts in 10,000. Fig 2.7 Slight dip in E(r) or J(r) moving from surface to center for a round wire in the low frequency limit. In this case radius a = 20 and skin depth δ = 200. (c) Review of the round wire solution To conclude this section, we state in full notation the solution of the round wire as outlined above, using ω rather than δ as the argument of interest, where recall δ = 2/ωμσ so z = 2 (r/δ) = r ωμσ . The following two expressions are (2.2.22) and (2.2.25) with ( β/jω) = ejπ/4ωμσ /ω as in (2.2.12): Chapter 2: The Round Wire and the Skin Effect 81 E(r,ω) = E(a,ω) ber0[r ωμσ ] + j bei 0[r ωμσ ] ber0[a ωμσ ] + j bei 0[a ωμσ ] (2.3.11) B(r,ω) = E(a,ω) ber 1[r ωμσ ] + j bei 1[r ωμσ ] ber0[a ωμσ ] + j bei 0[a ωμσ ] (-ejπ/4μσ ω ) (2.3.12) The ratio is then B(r,ω) E(r,ω) = ber1[r ωμσ ] + j bei 1[r ωμσ ] ber0[r ωμσ ] + j bei 0[r ωμσ ] (-ejπ/4μσ ω ) . (2.3.13) The time-domain fields are, from (2.1.1) a nd our assumptions (2.2.4) (d) and (e), E(x,y,z,t) = ej(ωt-βdz) E(r,ω ) z^ B(x,y,z,t) = ej(ωt-βdz) B(r,ω ) θ^ ( 2 . 3 . 1 4 ) so that, in terms of the complex value E(a, ω), E(x,y,z,t) = ej(ωt-βdz) E(a,ω) ber0[r ωμσ ] + j bei 0[r ωμσ ] ber0[a ωμσ ] + j bei 0[a ωμσ ] z^ ( 2 . 3 . 1 5 ) B(x,y,z,t) = - e j(ωt-βdz) E(a,ω) ejπ/4μσ ω ber1[r ωμσ ] + j bei 1[r ωμσ ] ber0[a ωμσ ] + j bei 0[a ωμσ ] θ^ . In the notation of Section 1.6 (d) these equations can be written E(x,y,z,t) = ej(ωt-βdz) ejφez(r,ω) E(r,ω) z^ E(r, ω) = ejφez(r,ω) E(r,ω) B(x,y,z,t) = ej(ωt-βdz) ejφbθ(r,ω) B(r,ω) θ^ B(r,ω) = ejφbθ(r,ω) B(r,ω) (2.3.16) where E(r,ω) = |E(r,ω)| and B(r,ω) = |B(r,ω)| are real. As shown in (1.6.6), the physical fields could be taken as either of the following pairs E phys(x,y,z,t) = Re{ E(x,y,z,t) } = cos[ ωt - βdz + φez(r,ω) ] E(r,ω) z^ Bphys(x,y,z,t) = Re{ B(x,y,z,t) } = cos[ ωt - βdz + φbθ(r,ω) ] B(r,ω) θ^ (2.3.17) or E phys(x,y,z,t) = Im{ E(x,y,z,t) } = sin[ ωt - βdz + φez(r,ω) ] E(r,ω) z^ Bphys(x,y,z,t) = Im{ B(x,y,z,t) } = sin[ ωt - βdz + φbθ(r,ω) ] B(r,ω) θ^ (2.3.18) Chapter 2: The Round Wire and the Skin Effect 82 (d) Plots of the round wire solution for Belden 8281 at 5 MHz. Here is so me Maple code to generate va rious plots, where we arbitrarily set E 0 = E(a,ω) = 1 volt/m, μ = μ0 = 4π x 10-7, σcopper = 5.81*107, ω = 2π [ 5 MHz ], and a = 394 μ -- all as appropriate for the center conductor of Belden 8281 coaxial cable. Notice that the factor μσ ω = 4π*5.81/2π 106 = 10-3 2*5.81 = 3.4 x 10-3 causes B to be small even at the surface r = a. We first set in the parameters just quoted, and then do the plots as follows, using (2.3.11) for E and (2.3.12) for B ( j = I in Maple) Fig 2.8 E = Magnitude of E φez = Phase of E Chapter 2: The Round Wire and the Skin Effect 83 Fig 2.9 B = Magnitude of B φbθ = Phase of B Fig 2.10 B/E = Magnitude of B/E φ bθ-φez = Phase of B/E Regarding the fast cycling of the phases of E and B, recall the discussion above (2.1.7) concerning the notion of the field being a highly damped radial wave, and below (2.1.7) where it was noted that in the 1D analog, the amplitude drops to 1/e when that radial wave has progressed a mere π/2 worth of phase. We see that happening here for both E and B. The nature of these plots for moderate to large z = r ωμσ can be obtained from the large z limit of the Jν functions as noted earlier, Jν(ej3π/4z) = M ν(z) ejθν(z) . z = r ωμσ (2.3.3) Mν(z) ≈ exp(z/ 2 ) 2πz θν(z) ≈ (z/ 2 ) + (π/2) [ ν - 1/4 ] . (2.3.5) Example : For the electric field in (2.3.16) we have this large z limit, E0 J0(ej3π/4r ωμσ ) J0(ej3π/4a ωμσ ) = E0 M0(r ωμσ ) M0(a ωμσ ) ejrωμσ/2 ejaωμσ/2 ≈ E0 exp(z/ 2 ) exp(za2 ) 2πza 2πz ej(r-a) ωμσ/2 Chapter 2: The Round Wire and the Skin Effect 84 ≈ E0 exp[- (a-r) ωμσ/2 ] a r e-j(a-r) ωμσ/2 ≈ E0 exp[- (a-r)/ δ] a r e-j(a-r)/ δ δ ≡ 2/(ωμσ) which shows both the exponential decay in magnitude and the phase linear in r, φez(r,ω) ≈ -(a-r)/δ . Again, this is reminiscent of the 1D skin depth solution shown in (2.1.8). E(x,ω) = E(0,ω ) e-x/δ e-jx/δ . (2.1.8) 2.4 The Surface Impedance Z s(ω) of a Round Wire A piece of round wire can be thought of as a resistor. Consider Fig. 2.11: Fig 2.11 Here a piece of finite- σ wire is attached to a pair of σ = ∞ contacts. The total impedance of the wire is then determined by Z = V/I ohms where V is the volta ge applied to the contacts and I is the total current through the wire. Alternatively, one could probe the wire along its su rface as shown by the two arrows separated by dz. There is some voltage dV between the probes due to the field E z(a) ≡ E(a) at the surface of the wire. By definition, the surface impedance per unit length is Z s ≡ (- dV/dz)/I = E(a) / I ohms/m . (2.4.1) Since the fields and currents derived under th e assumptions (2.2.4) vary only with r, Z s is independent of z and we get V = V(0) - V(L) = - ∫0 L dV = - ∫0 L (dV/dz) dz = ∫0 L I Zs dz = I Z s L = I Z (2.4.2) so Z = Z s L . ( 2 . 4 . 3 ) Chapter 2: The Round Wire and the Skin Effect 85 Our analysis above treats the infinitely long wire, so one must imagine L here as very large compared to the wire radius a, so that end effects influencing Z s can be ignored. As one might expect, Z s plays a role in transmission line attenuation. (a) Expressions for Surface Impedance To com pute the surface impedance of the round wire, we have to make a connection to the total current I in the wire. This time, our "math loop" is a circular ring lying just below the wire surface as shown in red in Fig 2.11. Apply (2.2.1) to this loop (with ε = 0) to get: 2πaB(a) = μI . (2.4.4) Thus, from the Z s definition (2.4.1), Z s = E(a)/I = E(a) μ/[2πaB(a)] = ( μ/2πa) E(a)/B(a) . (2.4.5) Recalling from (2.2.24) that E(a) = - (j ω/β) J 0(βa) J1(βa) B ( a ) . (2.2.24) we find that Zs = Zs(ω) = - (μ/2πa) (jω/β) J0(βa) J1(βa) or Z s(ω) = -jωμ 2πaβ J0(βa) J1(βa) ( 2 . 4 . 6 ) where β = ( 2 /δ) ej3π/4 and δ ≡ 2/ωμσ as in box (2.2.30). Using these last two facts and the fact that ej3π/4 is a square root of -j, the leading factor may be written -jωμ 2πaβ = -j2/(σδ2) 2πa ej3π/4 (2 /δ) = ej3π/4 2 πaσδ giving this alternate form for (2.4.6) in which ω does not explicitly appear, Zs(ω) = ej3π/4 2 πaσδ J0(βa) J1(βa) β = ( 2 /δ) ej3π/4 = (j-1)/δ (2.4.7) Below we shall use form (2.4.7) to plot Z s(ω) as a function of skin depth δ. Chapter 2: The Round Wire and the Skin Effect 86 Equation (2.4.7) is, as expected, rather complex. In terms of the Kelvin functions defined in (2.3.1) we may write (2.4.6) as Zs(ω) = -jωμ 2πa (2/δ) ej3π/4 ber0[2(a/δ)] + j bei 0[2(a/δ)] ber1[2(a/δ )] + j bei 1[2(a/δ)] . (2.4.8) According to (2.3.1) one finds, with α ≡ ej3π/4, that ber ν'(z) + j bei ν'(z) = dJν(αz) dz = dJν(αz) d(αz) d(αz) dz = αJν'(αz) = ej3π/4 Jν'(ej3π/4z) . (2.4.9) Then since J 0'(x) = -J1(x) one gets ber 0'(z) + j bei 0'(z) = ej3π/4J0'(ej3π/4z) = - ej3π/4 J1(ej3π/4z) = - e j3π/4 [ber1(z) + j bei 1(z)] . (2.4.10) Then (2.4.8) may be rewritten as Zs(ω) = +jωμ 2πa(2/δ) ber0[2(a/δ)] + j bei 0[2(a/δ)] ber0'[2(a/δ)] + j bei 0'[2(a/δ)] (2.4.11) and this form for Z s(ω) appears in Matick p 104 (4-28). (b) Low frequency limit of Z s(ω) Small ω => large δ => small β, so we expand both Bessel functions of (2.4.7) for small argument: [ Spiegel 24.5 and 24.6 ] J 0(x) ≈ 1 - x2/4 J1(x) ≈ (x/2)(1 - x2/8) ⇒ 1/J1(x) ≈ (2/x) (1 + x2/8) => J 0(x)/J1(x) ≈ (2/x) (1 + x2/8) (1 - x2/4) ≈ (2/x)(1-x2/8) = 2/x - x/4 => J 0(βa)/J1(βa) ≈ 2/(βa) - (β a)/4 . Then from (2.4.6) Z s(ω) = -jωμ 2πaβ J0(βa) J1(βa) ≈ -jωμ 2πaβ [2/(βa) - (βa)/4 ] = -jωμ πa2β2 + jωμ 8π = -jωμ πa2[-jωμσ] + jωμ 8π // β2 from (2.2.3) or Chapter 2: The Round Wire and the Skin Effect 87 Zs(ω) = 1 σπa2 + jω μ 8π = Rs + jωLs // low frequency limit (2.4.12) The first term is the uniform DC resistance of the wire per unit length, normally written ρ/A as in (C.1.1) of Appendix C. The second term is j ω times the DC internal inductance L i = (μ/8π) H/m, as derived in (C.3.10). Recall that this is exactly 50 nH/m if μ=μ0, quite small, and inde pendent of radius. (c) High frequency limit of Z s(ω) We first use (2.3.3) to write (2.4.6) as Z s(ω) = (-jωμ/2πaβ) [ M0(2 a/δ) / M1(2 a/δ) ] exp[ j{ θ0(2 a/δ) - θ1(2 a/δ)}] . (2.4.13) Since large ω ⇒ small δ ⇒ large arguments for the functions in (2.4.13), we use these large z limits which can easily be obtained from (2.3.5) using ν = 0 and 1, M0(z) / M1(z) = [ 1 + 1 22 z + O(1/z2) ] θ 0(z) - θ1(z) = - [ (π/2) + 1 22 z + O(1/z2) ] . (2.4.14) Insertion of these large-argument expressions into (2.4.13) with z = 2 δ/a gives Zs(ω) = (-jωμ/2πaβ) (1 + 1 22 2 a/δ ) exp(-j [ π/2 + 1 22 2 a/δ ]) = (-jωμ /2πaβ) [ 1 + δ /(4a) ] exp(-j [ π/2 + δ /(4a) ]) = e-jπ/2ωμ 2πa(2 /δ) ej3π/4 [ 1 + δ /(4a) ] e-jπ/2 e-jδ/(4a) . // -j = e-jπ/2 The phasor factors combine to give e -jπ/2 e-j3π/4 e-jπ/2 = e-jπ[1+3/4] = e-jπ[2-1/4] = e-jπ2 ejπ/4 = ejπ/4 = (1+j)/ 2 and then Z s(ω) = ωμ 4π(a/δ) (1+j) [ 1 + δ/(4a) ] e-jδ/(4a) . (2.4.15) Then if δ << 4a the last two factors are unity and we have Chapter 2: The Round Wire and the Skin Effect 88 Zs(ω) ≈ ωμ 4π(a/δ) (1+j) = ωμδ2 4πaδ (1+j) = ωμ[2/ωμσ] 4πaδ (1+j) // using δ2 = 2/ωμσ from (2.2.20) ≈ 1 σ(2πa)δ (1+j) δ << 4a . (2.4.16) Writing this as the sum of a resistive and inductive part, Zs(ω) = Rs(ω) + jω Ls(ω) ( 2 . 4 . 1 7 ) we find R s(ω) = 1 σ(2πa)δ = ω Ls(ω) = XLs(ω) . (2.4.18) The inductance can be written several ways, L s(ω) = 1 ωσ(2πa)δ = 1 ωσ(2πa)2/ωμσ = 1 2πa μ 2σω . (2.4.19) The resistance has a simple interpretation. It is R = 1/( σA) where area A = (2 πa)δ . This is the area of a thin washer at the periphery of the wire of thickness δ . The inductance is harder to understand. Its origin can be traced back to (2.4.5) above which shows that the phase of Z s is equal to the phase of the ratio E(a)/B(a). It is a result of Maxwell's curl equations that the phase of this ratio as seen in (2.4.16) is π/4 at the surface of a conducto r in the skin effect limit. The inductive reactance is the same as the resistance, but the inductance itself increases as frequency decreases, behaving as L ~ 1/ ω as shown. Quantity R s(ω) in (2.4.18) is called R hf by Matick p 105 in his (4-35), and (2.4.16) appears as (4-36). (d) Plots of Z s(ω) versus skin depth δ From (2.4.7) we found that Z s(ω) = ej3π/4 2 πaσδ J0(βa) J1(βa) β = ( 2 /δ) ej3π/4 = (j-1)/δ δ ≡ 2/ωμσ . (2.4.7) This is in SI units, but we will use a = [a( μ)10-6] m and δ = [δ(μ) 10-6] m and σ = 5.81 x 107mho/m for copper, where a( μ) and δ(μ) means the wire radius and skin depth in microns. Then: Zs(ω) = Nej3π/4 2 πa(μ)σδ(μ) J0(βa) J1(βa) ohms/m N = 1012 The two limits obtained above were : Chapter 2: The Round Wire and the Skin Effect 89 Zs(ω) ≈ 1 σπa2 + jω μ 8π = 1 σπa2 + j 1 4πσδ2 small ω, large δ (2.4.12) Real part goes to a constant, imaginary part decays as 1/ δ2 Zs(ω) ≈ 1 σ(2πa)δ (1+j) = 1 σ(2πa)δ + j 1 σ(2πa)δ large ω, small δ (2.4.16) Real and Imaginary part are the same and blow up as 1/ δ In our units above, the low frequency constant limit is R LF ≡ 1 σπa2 = N σπa(μ)2 . Example : For a = 1000 μ, the DC resistance is R LF = 1/(σπa2) = .00548 Ω/m. Since a = 1000/25.4 = 39.37 mils, 2a = 78.74 mils. From the following British units graphic, 12 gauge house wire has 2a = 80.808 mils and has R =.001588 Ω/ft which is .005210 Ω/m ) Here is Maple code which plots the real (red) and imaginary (black) part of ln Z s(ω) as a function of δ, and also computes the constant limit (gray) just mentioned. The copper wire radius is set to a=1000 μ. Chapter 2: The Round Wire and the Skin Effect 90 Fig 2.12 Plot of surface impedance ln Z s as function of skin depth δ ≈ 40 to 1000 μ for a copper wire of radius 1000 μ. Red is real part, black is imaginary. The general idea is that surface impedance goes up as δ goes down (left end of graph). Here is the same plot for δ = 100 to 1000 without the natural logs (ln = "log" in Maple) : Fig 2.13 Plot of surface impedance Z s as function of skin depth δ = 100 to 1000 μ for a copper wire of radius 1000 μ. Red is real part, black is imaginary. Either plot type realizes the two limits discussed above. Chapter 2: The Round Wire and the Skin Effect 91 For the limited range δ = 1 to 10 μ the above plot has this appearance ( the red and black curves are superposed and the gray constant line at .0055 Ω/m is indistinguishable from the x axis) : Fig 2.14 Plot of surface impedance Z s as function of skin depth δ = 1 to 10 μ for a copper wire of radius 1000 μ. Red is real part, black is imaginary. The resistance of our near-12-gauge house wire at δ = 1 micron (4.37 GHz) is about 2.7 Ω/m, which is about 500 times larger than the DC resistance. 2.5 Surface Impedance for a Transmission Line What is the surface i mpedance of an arbitrary conduc tor? As we have seen, a significant amount of work was needed to obtain the exact resu lt even for the simple geometry of a round wire with a symmetric current distribution. One can repeat this calculation fo r other geometries such as a parallel plate line. The general nature of the result is always the same when δ is much smaller than the depth of the conductor. That result is this (with comparison) Zs(ω) ≈ 1 σpδ (1 + j) // general case (2.5.1) Zs(ω) ≈ 1 σ(2πa)δ (1+j) δ << 4a . // round wire (2.4.16) where p is the effective perimeter distance around the cross-sectional surface of a conductor where significant current flows. For the round wire this was D = 2 πa, the circumference. For a parallel plate line of width w, D = w. Consider these two possible transmission line cross sections: Chapter 2: The Round Wire and the Skin Effect 92 wradius at b F i g 2 . 1 5 Cross section of a parallel plate line Cross section of a twin-lead In both cases we assume a frequency ω such that skin depth δ is small compared to the thickness of the conductors. Although the total cross s ectional perimeter of one of the parallel plates is 2w + 2t, it seems clear that the length of the "active surface" is only w, and one sets p = w in the surface impedance formula. For the twin lead case with leads assumed fa r apart (b >> a), both conductors are immersed in roughly uniform active fields, so the full p = 2π a is applicable. As the two round wires are brought very close togeth er, certainly there will develop an asymmetry so that the currents are largest on the parts of the wires closest to the other wire. In this case, one must make an estimate of the "effective perimeter" p . Here is a picture, Fat twinlead Fig 2.16 where we have indicated a crude graphical estimate of the "active perimeter" of current flow. King [p 30 Eq (45)] quotes an approximate surface impedance result for the case of Figure 2.16. The effective distance is, p = 2πa 1 - (2a/b)2 . a = wire radius, b = center line separation (2.5.2) If the gap between the conductors is a/6, a rough estimat e for Fig 2.16, then the radical in this formula becomes .38, so the dark lines shown should cover 38% of the circumference. If the conductors almost touch, then p becomes extremely small. King makes the interesting remark (p 30) that "accurate formulas for the internal (i.e., surface) impedance of one cylindrical conductor in the presence of another with different radius are not available." From the potential results of Chapter 6 below one can obtain the surface charge density n( θ) in terms of ∂ nφ on such cylindrical conductors and thus the charge partial wave moments ηm used in Appendix D. From these one could find E z at the conductor surfaces using (D.4.9) and that would seem to determine Zs. In general, the high frequency skin current will be large where the E and B fields are large. These fields are large where the electric field would be large in a capacitor whose "plates" are the two conductors in cross section. Recall that such a 2D capacitor problem seeks potential φ as a solution of the Chapter 2: The Round Wire and the Skin Effect 93 equation ∇2 2Dφ(x,y) = 0. In regions where φ is very large, E = - ∇φ will also be large. This subject is addressed in Chapter 5 below. There is an interesting transmission line "parad igm shift" which occurs as one moves from the low frequency domain to that of high frequency. For small ω, one thinks of the currents in the two conductors of a transmission line as being there because they are "a pplied" by some external agency. The current then creates a B field around each wire which, si nce it is changing, creates an E field. In the high frequency skin-effect limit, it is easier to think of the currents in the conductor surfaces as being generated by the field activity near the surfac es. The E and B fields just outside the conductors force themselves slightly into the su rface. The resulting E field in the surface layer is then what creates the current. Matick's Chapter 4 computes the surface impedance for the round wire and for strip line conductors (but not the official Stripline). Chapter 3: Transmission Line Preliminaries 94 Chapter 3: Transmission Line Preliminaries 3.1 Why is there no free charge inside a conductor or a dielectric? I magine that at time t = 0 there were some free charge ρ inside a medium having conductivity σ. What would this free charge do? Intuition suggests that the individual charges in the little charge cloud would repel each other and the cloud would spread out until it encountered boundaries. In this Section we put that intuition on a more technical footing. As discussed in Appendix E, for a non-neutral medium Ohm's Law takes the form J = σE - D grad ρ , ( 3 . 1 . 1 ) where J is conduction current and the second term, associat ed with Fick's Law, is non-zero when the free charge density ρ is non-zero. This second term is a diffusion term, D is the (electron) diffusion constant for the medium at hand, and the diffusion current flows from a region of high charge density to one of lower density, hence the minus sign. Taking the divergence of the above equation, one finds div J = σ div E - D ∇ 2 ρ or -∂ tρ = σ ρ/ε - D ∇2 ρ // using (1.1.25) for div J , and (1.1.3) with (1.1.6) for div E or ∂ tρ - D ∇2 ρ + (σ/ε)ρ = 0 ( 3 . 1 . 2 ) or ∂ tρ - a∇2ρ - bρ = 0 where a ≡ D, b ≡ - (σ/ε) . (3.1.3) Now let ρ ' = ρ e-bt be an "adjusted" charge density. Then, since ρ = ρ'ebt, (3.1.3) becomes [(∂tρ')ebt + ρ'bebt] - a ebt∇2ρ' - bebtρ' = 0 or ∂ tρ' - a∇2ρ' = 0 ( 3 . 1 . 4 ) which is the standard heat/diffusion equation. If one starts at t = 0 with a point charge ρ' = q δ (r) at the origin, and if one assumes an infinite isotropic medium, one finds that at time t the charge density is given by ρ'(r,t) = q exp(-r 2/4at) / (4πat)3/2 . t ≥ 0 (3.1.5) This is the 3D causal free-space propagator (Green func tion) for the heat equation. It is the solution of (∂t - a ∇2) ρ'(r,t) = δ (r)δ(t) ρ'(r,t) = 0 for t<0 . (3.1.6) See Stakgold (5.133) and (5.136). In n spatial dimens ions, the propagator is as in (3.1.5) with 3/2 → n/2 and is derived in the text leading up to Stakgold (5.140). Therefore, if we consider (3.1.3) for charge density ρ Chapter 3: Transmission Line Preliminaries 95 (∂t - a∇2 - b)ρ(r,t) = δ (r)δ(t) (3.1.7) replacing ρ = ρ'ebt gives ∂tρ' - a∇2ρ' = δ (r)δ(t)e-bt = δ(r)δ(t) which is the same as (3.1.6). Ther efore, the solution of (3.1.7) is ρ(r,t) = ρ 'ebt = q ebtexp(-r2/4at) / (4πat)3/2 ρ(r,0) = q δ(r) or ρ(r,t) = q e -(σ/ε)texp(-r2/4Dt) / (4πDt)3/2 ρ(r,0) = q δ(r) . (3.1.8) The first factor e -(σ/ε)t says that ρ (r,t) decays exponentially in tim e in a uniform manner over space, while the second term says that the rough radius of the diffusing charge cloud is given by r = 4Dt . The main point of all this math is the following: if there is any free charge in a medium, it goes away in a timely manner. In our idealized an alysis above, it runs off to r = ∞, but in a finite medium it runs off to the boundary surface of the medi um and becomes surface charge. Let's now look at two extreme cases. For a good conductor with a low diffusion rate, equation (3.1.2) becomes ∂ tρ + (σ/ε)ρ = 0 ( 3 . 1 . 9 ) which has the obvious solution ρ(r,t) = ρ(r,0) e -(σ/ε)t which replicates the first factor of (3.1.8). The charge just "flows away" due to the large σ . For a dielectric with a very small conductivity, equation (3.1.2) instead becomes D ∇ 2ρ - ∂tρ = 0 ( 3 . 1 . 1 0 ) which is just the heat equation whose im pulse response solution is (3.1.8) with σ = 0, as was shown in (3.1.5). In this case, the charge at least has time to diffuse out before it goes away! There are then two time constants involved. The first is for the e-(σ/ε)t factor where τ = ε/σ. We can estimate this time constant for a conductor and dielectric using ε ≈ ε0, and copper σ = 5.81 x 10 7 mho/m ε0 = 8.8541877 x 10-12 farad/m (from 1.1.28) τ = ε/σ ≈ 10 -11 / 108 ≈ 10-18 sec (3.1.11) so in copper, free charge runs off to the surface in one thousandth of a femtosecond, so we don't worry about the diffusion time constant. Chapter 3: Transmission Line Preliminaries 96 For a dielectric with σ = 10-15 mho/m we get τ larger by 1023 which is then 105 seconds or about a day. But in this case, the diffusion mechanism wins out. As an example, for pure silicon, D ≈ 40 cm2/sec = 4x10-3 m2/sec. The time to diffuse from a delta function out to say r = 1 mm is given by r = 4Dt = t = r2/(4D) = (10-3m)2 / (4*10-3 m2/sec) = (1/4) x 10-3 sec (3.1.12) so in this case the charge is pretty much gone in a quarter of a millisecond. We arrive then at this fact: Fact 1 : In a transmission line, charge exists only on the surface of conductors. (3.1.13) Comment : If one wants an initial charge distribution ρ '(r,0) to be something other than a delta function, one may use this solution to the heat equation (3.1.4), ρ'(r,t) = [2r πat ]-1 ∫0 ∞ r'dr' ρ '(r',0) { exp[-(r-r')2/4at] - exp[-(r+r')2/4at] } (3.1.14) which appears in Polyanin 1.2.3-10. Setting ρ'(r',0) = q δ (r') = q δ(r')/(4πr'2) then replicates the earlier result (3.1.5), after using L'Ho ^pital's Rule on the integrand. The reason δ (r) = δ (r)/4πr2 is that it makes ∫dV δ(r) = 1 when integrated over a sphere of any radius. There is an exception to our rule that ρ = 0 inside a conductor. If magnetic fields are present in the right manner, it is possible to have an extremely small ρ ≠ 0 inside a conductor -- so small that one can ignore it in any practical application. An example of this situ ation is presented in Section N.7 which concerns what we call the Radial Hall Effect in a round wire. The tiny charge density is required to produce a radial Hall field which offsets conduction electrons' radial Lorentz force. In the regular Hall effect, this Hall field is generated by surface charges on opposite faces of a sa mple, but in the round wire case only one surface is available, which is the surface of the round wire. 3.2 How thick is the surface charge layer on a conductor? This is a fascinating subject and the i nterested reader will find an analysis in Appendix E from which we now quote. Since there is no free charge in the dielectric outside the conductor, and since electrons at normal temperatures cannot jump off the conductor due to its so-called work function, the surface charge is actually a layer just below the nominal surface of the conductor, but can essentially be regarded as being right at the surface. The situation is much diffe rent when a conductor is immersed in a solution of charge-carrying ions or molecules. It turns out that the charge density decays expone ntially away from the surface into the conductor and drops to 1/e of its surface value at a distance called the Debye length. For copper, this distance is roughly 0.6A (Angstroms), which is 6 x 10 -11 m. The crystal spacing for copper is 3.6A, and the copper atom radius is about 1.3A. Thus, Chapter 3: Transmission Line Preliminaries 97 Fact 2 : The thickness of the surface charge density on the surface of a conductor is incredibly small. For copper, it is less than the radius of one copper atom, and the general result applies to any metal. (3.2.1) Table (2.3.9) shows that the skin depth δ for copper at 100 GHz is about 0.2 microns which is 2x10 -7 m ≈ 2000A. Even at this huge frequency, the skin depth is still about 4000 times larger than the thickness of the surface charge layer. At 1 GHz this ratio is 40,000. Fact 3 : Whereas current can exist "deep" under the surface of a conductor, even when the skin effect is dominant, the surface charge can always be thought of as being exactly at the surface. (3.2.2) Motions of surface charges can create a 2D current density on the surface, which we might refer to as a Debye Surface Current (see Section D.9). In a transmi ssion line, even in the extreme skin effect regime, any such Debye surface current is completely swamped by the skin effect current, and so can be ignored. In effect, one can regard such a Debye current as a ve ry tiny fraction of the total skin effect current. We just saw above how the Debye current layer might be 0.6 A thick, while the skin effect current at 100 GHz is 2000A thick. 3.3 How does loss tangent affect dielectric conductivity? The total current in a dielectric may be written, as noted in (2.2.1), Jtot = jωεdE + σdE . ( 3 . 3 . 1 ) The first term is the displacement current and the second term is the conduction current. At high frequencies (say 1 GHz), the dielectric constant ε acquires a small imaginary part due to the presence of absorption resonances in the medium at infrared frequencies. One can write, εd = ε'd - jε"d = ε'd [ 1 - j (ε"d/ε'd) ] = ε'd [1 - j tan L] , tan L ≡ (ε"d/ε'd) . (3.3.2) If one plots εd in the complex plane, tan L (called the loss tangent , aka tanδ) is the tangent of the small angle θL of the triangle whose perpendicular sides have length ε'd and ε"d where ε"d is normally very small. That is to say, the loss tangent is (minus) the ratio of the small imaginary part to the dominant real part of εd. tanL is commonly referred to as the dissipation factor . When this expression is inserted into (3.3.1) the result is, Jtot = jωεd E + σdE = jωε'd [1 - j tan L]E + σdE = jωε 'd E + ( σd + ωε'd tanL) E = jωε'd E + σeff E . ( 3 . 3 . 3 ) In effect, the dielectric has now acquired an effective conductivity, σ eff = ( σd + ωε'd tanL) . ( 3 . 3 . 4 ) Chapter 3: Transmission Line Preliminaries 98 Because the DC conductivity of a good dielectric is so small, the loss tangent contribution to σeff dominates even at quite low frequencies. For polyethylene we can use these ballpark numbers, σd ≈ 10-15 mho/m ε'd ≈ 2.3 tan L ≈ 2 x 10-4 (3.3.5) taken from the following 2008 studies of Low and High Density Polyethylene done by Eaton and Kmiec, and for conductivity, Fig 3.1 http://www.sdplastics.com/polyeth.html San Diego Plastics, Inc. Note : This table claims ρd = 1015 ohm-cm = 1013 ohm-m, but most other sources give larger values. We assume ρd ~ 1017 ohm-cm = 1015 ohm-m and therefore σd ~ 10-15 mho/m. It does not matter much! Even at 1 Hz, the loss tangent contribution dominates in (3.3.4). Using the above figure for tan L, here are a few values of σeff ≈ (εd tanL)ω versus frequency: (f = .0 is really 100 = 1 Hz) Chapter 3: Transmission Line Preliminaries 99 // = 2.6 x 10-3 so σ d,eff(ω) ≈ (εd tanL)ω = 4.1 x 10-15 ω = 2.6 x 10-5 f(GHz) // PE (3.3.6) Thus, in a transmission line, although the nominal DC dielectric conductance might be 10 -15 mho/m, at operating frequencies the effective σ d is much larger, being for example 2.6 x 10-5 mho/m in polyethylene at 1 GHz. However, even if we replace σd by the much larger σeff in the complex dielectric constant ξ d, ξd = ε'd + σeff/jω = ε'd + [σd + ωε'd tanL]/jω ≈ ε'd + [ωε'd tanL]/jω = ε'd [1 + tan L/j] ≈ ε'd ≈ εd ( 3 . 3 . 7 ) we still find for a material like polyethylene that ξ d ≈ εd ( 3 . 3 . 8 ) at least to 2.5 GHz. It is not hard to write an expression for tan L as a function of frequency since it involves the real and imaginary parts of the dielectric constant εd(ω) which has infrared resonances dependent on the medium. Since RF frequencies up to perhaps 10 GHz are much less than infrared frequencies, although tan L does increase somewhat with ω in this range, one still has tan L << 1. Advanced dielectrics typically have tan L in the range .002 or less at 10 GHz. 3.4 Size of E fields in conductor and dielectri c; con servation of total current at a boundary We know from (1.1.48) that the following E field condition applies at a bound ary between two media, where n refers to the normal component, ξ1En1 = ξ2En2 // frequency domain (1.1.48) or (ε 1 + σ1/jω) En1 = (ε2 + σ2/jω) En2 . (3.4.1) Let 1 = dielectric and 2 = conductor. From (3.3.8) we set ξ 1 ≈ ε1 (at least for f < 10GHz), and from (2.2.3) we set ξ2 ≈ σ2 /jω for f << 109 GHz (for polyethylene and copper), so (3.4.1) then reads Chapter 3: Transmission Line Preliminaries 100 ε1 En1 ≈ (σ2/jω) En2 => ratio = ⎪⎪⎪ ⎪⎪⎪En1 En2 ≈ σ2 ωε1 . (3.4.2) We can look at some typical numbers, ε 1 = 2.3 ε0 ( p o l y e t h y l e n e ) ( 3 . 4 . 3 ) σ2 = 5.81 x 107 mho/m (copper) ε0 = 8.85 x 10-12 farad/m so σ2 ωε1 = 5.81 x 107 2πf 2.3 8.85 x 10-12 ≈ (1/2)1018/f ≈ 109/[2f(GHz)] (3.4.4) At a high frequency of f ≈ 500GHz the ratio in (3.4.2) is ~ 106, and at lower frequencies the ratio only increases. Thus, we arrive at these useful facts: Fact 1: The total current in a dielectric is dominated by displacement current, while that in a conductor is dominated by conduction current. (3.4.5) Fact 2 : At a boundary between a good dielectric and a good conductor, the normal E field is at least 1 million times larger in the dielectric than it is in the conductor for frequencies under 500 GHz. (3.4.6) Fact 3: This large jump in E n at the boundary must be supported by a significant surface charge density n on the boundary since, according to (1.1.47), n = ε1En1 - ε2En2 ≈ ε1En1. (3.4.7) Imagine now a tiny patch of area (bordered in red) on the surface between a conductor and a dielectric, Fig 3.2 Defining a total current J tot,n ≡ jωεEn + σEn, as in (2.2.1), we have show n that this total current flows right through the area patch but changes its nature fr om mostly conduction current on one side to mostly displacement current on the other side. In the next secti on, we identify the normal direction with the local radial direction. Then the total current passing thr ough a tiny square patch like that in Fig 3.2 can be regarded as being "fed" by the radial current J r just inside the conductor where J r = σEr. This current feeds the surface charge on the boundary which in turn creates a large E field and thus a large displacement current in the dielectric. We sometimes refer to this mechanism as "charge pumping". Chapter 3: Transmission Line Preliminaries 101 3.5 The TEM mode fields and currents for an ideal transmis sion line In this and the next section, we take a crude qua litative look and the various E,B and J components first for an ideal transmission line, then for a practical one. An example is repeatedly used in which the conductor of interest is the round center conductor (radius a = 1 mm) of a properly terminated 75 Ω coaxial cable driven by 7.5 volts, and thus having a current of 100 mA. The two tables obtained (one ideal, one practical) mainly serve as an exercise in a pplying the various concepts reviewed in previous sections. By "ideal" we mean that the conductors have ne ar infinite conductivity and the dielectric has zero conductivity. Consider a cross sectional view of one conductor of a transmission line having arbitrarily shaped conductors (the shape is uniform in the z direction). For a given point on the surface, define a cylindrical coordinate system (axis through red dot) so that r = radial direction = the normal outward from the surface (local x) θ = azimuthal direction = tangential to the surface in the cross section plane (local y) z = tangential to the surface along the transmission line (local and global z) Fig 3.3 The following table shows the qualitative sizes of va rious components of E,B and J (conduction current) near the surface of a transmission line conductor. Several regions of space are of interest: 1. Deep in the conductor, under the surface ch arge layer and under any current layer. 2. In the conductor, just under the surface charge layer, and in the skin current layer. 3. In the dielectric, just outside the super-thin surface charge layer. The reader is warned that the rest of this section and Section 3.6 make very tedious reading because an argument must be made for the general size of every single item in the two large Tables. We recommend that the reader just peruse the two Tables and ignore the Explanation sections unless there is an interest in some particular table value. Chapter 3: Transmission Line Preliminaries 102 First is the Table for the ideal transmission line conductor : Table 1: E,B,J for an ideal transmission line Region 1. Deep in the conductor, under the su rface charge layer and under any current layer. E r = 0 B r = 0 J r = 0 E θ = 0 B θ = 0 J θ = 0 E z = 0 B z = 0 J z = 0 Region 2. In the conductor, just under the su rface charge layer, and in the current layer. E r = small B r = 0 J r = small E θ = 0 B θ = large J θ = 0 E z = small B z = small J z = very large Region 3. In the dielectric, just outside the super-thin surface charge layer (explanations below): E r = large B r = 0 J r = 0 E θ = 0 B θ = large J θ = 0 E z = small B z = small J z = 0 (3.5.1) Explanations of Table Entries Region 1 : (the interior) In the interior we know that E must satisfy the Helmholtz equation (2.1.6a). Due to the powerful exponential effect of this equation (see (2.1.8) and (2.3.7) for the round wire), we know that E fields cannot exist deep inside the conductor, a nd can exist only in the skin depth region. Maxwell (1.1.2) says curl E = -jωB in the ω domain, so if E = 0 in the interior, so also is B. A "perfect conductor" has σ = extremely large, and δ = extremely small since δ = 2/μσω . Thus, conductor E and B fields can only exist very close to the surface. In region 1 of the above table, we show all fields as being 0 underneath the very thin current sheath. Since E = 0 in the perfect conductor interior, it follows from J = σE that J = 0 there as well (region 1). Thus, all current is c onfined to the thin current sheath of regions 2. Everything is quiet in Region 1. Region 2 : (the current sheath) As just noted, all current s flow in a very thin sheath at the surface of thickness δ. Since the thickness is tiny, the current density J z there is "very large" as marked in the table. Imagine a total current I flowing down the conductor, but it is restricted to flow only in the thin sheath. In this thin layer, there is some radial pumping of charge to the surface to "feed" the surface charge which is always changing in time, so we indicate a small J r term. As noted in Section 3.4, this same J r is "feeding" the total current flow through the surface, and the surface converts this total current from conduction current on the inside to displacement current on the outside. An argument will given below for why Jr is small compared with J z and we duly mark J r as "small" in region 2. Application of Ampere's Law (1.1.37) to the small red loop in Fig 3.3 (B θ = 0 on the left long edge) shows that the large J z sheath current creates a "large" B θ field in the sheath which grows from 0 on the sheath's inner boundary to some large value at the conductor surface. Ignoring dramatic μ differences, this Chapter 3: Transmission Line Preliminaries 103 Bθ then exists just outside the surface as we ll according to (1.1.42). We thus mark B θ as "large" in both regions 2 and 3. If I = 100 mA and a = 1 mm for a round conductor, then B θ = μ0I/(2πa) = 20 μ T at the wire surface. (Earth field is 32 μT). This is a large value for B θ in our current context. Since E = J /σ, even though J z is very large, σ is extremely large, so we shall mark E z as being "small". And since J r is already marked "small", we mark E r also as "small". The small radial current J r might create some small B z, so we throw in a small B z entry as well (see Region 3 below). The remaining three entries (B r, Eθ, Jθ) in region 2 we leave at 0, though they might have some very tiny values. Region 3: (the dielectric) Since we are now outside the surface charge layer, (1.1.47) says there is a large radial electric field E r which is supported by this charge de nsity (Gauss's Law), so we mark E r as "large" in region 3. The tangential electric fields are c ontinuous through the boundary according to (1.1.41). Therefore, we give E θ and Ez the same values they had in region 2. We already observed that B θ continues being "large" just above the surface. It was noted above that there is a radial pumping current J r inside the conductor. This pumps charge onto the conductor surface, and this J r is converted to displacement cu rrent in the dielectric as discussed above in Section 3.4 (think of a simple parallel plate capacitor where this also happens). This displacement current and J r are relatively small currents and they create a small B z field as we now crudely demonstrate. Consider a very tall and thin (small w) red math loop whose one edge lies parallel to the z direction between the conductors a nd whose top edge is very distant. Fig 3.4 Consider Ampere's law (1.1.37) rela tive to this loop and with respect to the displacement current flowing through the loop between the conductors, ∫{ H • ds = ∫S ∂tD • dS . (1.1.37) Integration of the small displacement current ∂tD passing through the loop gives some small non-zero value for the area integral on the right. Meanwhile , the line integral on the left has cancelling contributions from the vertical loop sides (w is very sm all), while the loop top is far away so contributes nothing. The result is some small H z and hence small B z in the region between the conductors. Since B z is a tangential field, it will exist also just inside th e conductor surface, as indicated by (1.1.42). Both these Bz fields are marked "small" in the table for regions 2 and 3. As a crude estimate, a loop of width w = λ/2 Chapter 3: Transmission Line Preliminaries 104 would capture a full I worth of displacement current, so our thin loop captures ~ I w/( λ/2). If I ~ 100 mA and λ ≈ 1 m, then Ampere's law above says (B z/μ0)w = I w/(λ /2) so Bz ≈ μ0 I(λ/2) = 4π x 10-7(0.1)(1/2) ≈ 6.3 x 10-8 = .06 μT, which is small compared to our 20 μT estimate for B θ. Since the dielectric has zero conductivity, the co nduction current components are all set to zero. In the dielectric, if we ignore the small E z and Bz field components relative to the large E r and B θ, we find that (see Fig 3.3) just outside the surface, the E and B fields are perpendicular and are both transverse to the z direction. Hence this is a TEM (Transverse Electric and Magnetic) mode of the transmission line. Their cross product is the Poynting vector (1/ μ) E x B which is in the +z direction coming at the viewer in Fig 3.3. This is the direction of power flow along the transmission line. (Jackson 6.109: S = E x H in SI units) The remaining two entries (B r, Eθ) in the region 3 we leave at 0, though they might have some very tiny values. 3.6 The TEM mode fields and currents for a pra ctical transmission line We now "turn on" the imperfections of the transmission line. As soon as σ in the conductor becomes large but finite, the infinitely thin current sheat h spreads out over some reasonable skin depth δ. At very low frequencies, the current J z is spread across the entire conductor and there is no Region 1. Fig 3.3 At higher ω there still is a Region 1, but we shall ignore it from now on. We are still interested in region 2 which is just below the surface charge layer. Recall from Section 3.2 that the surface charge layer remains nearly infinitely thin even for a non-perfect conductor. So here is the new table. The superscripts refer to descriptive sections below. Other values are just carried from the previous table. In order to make ba llpark magnitude estimates, we again assume that the transmission line is 75 ohms, is properly terminated, and is driven by a voltage of amplitude 7.5 volts, so the current is 100 mA. Chapter 3: Transmission Line Preliminaries 105 Table 2: E,B,J for a practical transmission line Region 2. In the conductor, just under the surf ace charge layer, and in the current layer. E r = small [c] B r = 0 J r = small E θ = 0 B θ = large J θ = 0 E z = small [a] B z = small J z = large [a] Region 3. In the dielectric, just outside the super-thin surface charge layer. E r = large [a] B r = 0 J r = small [b] E θ = 0 B θ = large J θ = 0 E z = small [a] B z = small Jz = leakage [b] (3.6.1) Explanations of Table Entries [a] Ez and Jz in the conductor; E z and Er outside the conductor Inside the conductor, a non-zero E z exists due to the current flow in the z direction and the finite conductivity of the conductor. As an estimate for a round wire not too close to the other conductor, assume that the wire has diameter 1 mm, and is operating at 1 GHz with a skin depth δ = 2 microns as in (2.3.9). The cross sectional area fo r current flow is then about 2 πrδ = 4π x 10-9 m2. If 100 mA flows through this wire, then J z = 0.1/(2πrδ) = 8 x 106 amps/m2, and this J z is marked "large" for region 2 in the above table. Then E z = Jz/σ = 8 x 106 / 5.81 x 107 = 0.14 volts/meter. This E z is marked "small" in the region 2 part of the above table. At lowe r frequencies where skin depth is larger, E z is less. Since E z is a tangential (parallel to c onductor surface) E field, according to (1.1.41) it has the same value in region 3, so that is also marked "small" above. In contrast, if the conductor separation is 0.5 cm, and if we crudely assume the E field is constant between the conductors, then E r between the conductors is 7.5 volts/ 5 x 10-3 m = 1500 volts/m. This is marked "large" in region 3 above. So in region 3 just outside the conductor, E r ~ 1500 V/m E z ~ 0.14 V/m ratio (E z/ Er) ≤ 10-4 (3.6.2) [b] Leakage: J r, Ez and Jz in the dielectric By "leakage" is meant conduction through the dielectric. As shown in (3.3.4), the effective conductivity in the dielectric is given by σ eff = ( σd + ωε'd tanL) . (3.3.4) For polyethylene, σd ~ 10-15 and can be ignored, while ε'd ≈ 2.3 ε0 and tanL ≈ 2x10-4 as in (3.3.5). For a frequency of 1 GHZ, we then find σeff ≈ ωε'd tanL ≈ 2π 109 * [2.3 * 8.85 x 10-12] * 2 x 10-4 ≈ 2.6 x 10-5 . (3.6.3) Chapter 3: Transmission Line Preliminaries 106 This is 12 orders of magnitude smaller than the σ of copper ~ 107, but it is 10 orders of magnitude larger than the DC conductivity of the dielectric ~ 10-15. To estimate the significance of this leakage at hi gh frequencies, we can compare the ratio of the leakage current to the displacement current in the dielectric (the currents flow through the same area so ratio is J leak/Jdisp) | I(leakage) I(displacement) | ≈ | (σd + ωεd tanL)E jωεd E | ≈ tanL ≈ 2 x 10-4 . (3.6.4) Thus, even at high frequencies, the effect of leakag e on the current flowing through the dielectric is quite small compared to the displacement current. The "radial" current J r has to support both the leakage current and the more significant displacement current, and we have just seen that the leakage part can be ignored. We carry region 3 "small" J r from the previous table since the leakage does not alter this fact. Finally, we already noted a small E z just outside the conductor, and si nce the dielectric has some very small leakage ( σeff), there will be some small J z in region 3 which we have marked "leakage". [c] Er and Jr inside the conductor We have already estimated that E r inside the conductor surface is less than 10-6 what it is outside the surface, see Section 3.4 Fact 2. Thus, if E r outside is 1500 volts/m as in our section (a) example, E r inside is less than 1.5 mV/m at 500 GHz, and is pr oportionally less than this at lower frequencies, so E r in region 2 is marked "small". In the example above we found E z ≈ .14 V/m inside the conductor. Thus we have Er << Ez inside the conductor which in turn means J r << Jz . Below we shall provide more support for the idea that J r << Jz . 3.7 The general shape of fields, charges, and currents on a transmission line (a) Eθ at a conductor surface vanishes We start by borrowing Fig B.6 from Appendix B (similar to Fig 3.3 above), Fig B.6 Chapter 3: Transmission Line Preliminaries 107 The figure shows a transmission line conductor of some arbitrary (but reasonably smooth) cross section shape. At the point of interest s we construct a cylindrical coordinate system as shown, such that the coordinates (r, θ,z) are appropriate for point s and its immediate neighborhood. Basically we approximate the piece of conductor surface near s as if it were the surface of a round wire of some radius r. At this point s, then, we can talk about fields E θ, Er, Bθ, and Br. For a transmission line we shall use Et to refer to the transverse components of an electric field, as opposed to the longitudinal component E z. In Cartesian coordinates Et = (Ex,Ey) and in the local cylindrical coordinates just defined at a surface point s, Et = (Er,Eθ). The important point is that E θ is our notation for the component of Et which at some surface point is tangent to the surface, while E r is normal to the surface ( we will also call this E n below). A fundamental assumption of transmission line theory is that the cross-section tangential electric field at a conductor surface vanishes, which is to say, E θ as defined above vanishes at all points on the surface. This assumption is examined in Appendix D.8 and here we accept it as fact. The basic idea is that surface charge is free to move along the conductor surf ace in a z=constant plane to neutralize any E θ that might develop, and this mechanism of maintaining E θ = 0 on the surface works from DC up to perhaps 1000 GHz. So: Fact 0: E θ = 0 at the surface of a transmission line conductor. (3.7.0) This assumption, stated in partial waves, appears in (D.2.27) and is one of tw o boundary conditions used in Appendix D to determine the internal fields of a round wire, the other boundary condition being (D.2.26). One can consider Fact 1 to be part of th e "electro-quasi-static" model of a transmission line. (b) The transverse vector potential components are small Fact 1: In the King gauge, for a trans mission line operating in the transmission line limit, the transverse vector potential is very small: | At| < 10-4 |Az| for f = 10 f c to 1000 GHz. (3.7.1) This is demonstrated in Appe ndix M, see (M.22). Frequency f c is a certain low end soft cutoff frequency that depends on the transmission line geometry. The basic idea is that in a transmission line the major current is in the z direction, and A ~ J according to the Helmholtz integral. Then since |J t| << |Jz|, one finds that |A t| << |Az| . (c) The scalar potential φ on a conduc tor surface By "conductor surface" we mean the boundary of a cross-sectional slice at z = constant through a transmission line conductor. In electrostatics one has E = - ∇φ and then E t = ∇tφ for the transverse electric field. In the neighborhood of a surface point s we write this as E θ = (1/r)∂θφ and Er = ∂rφ. Since Fact 0 says E θ = 0 at any s on the surface, we conclude that φ = constant all the way around the conductor boundary. This is fine for ω = 0, but for ω > 0 we have from (1.3.1) that E = -∇φ - jωA and so Et = -∇tφ - jωAt ( 3 . 7 . 2 ) Chapter 3: Transmission Line Preliminaries 108 and now it is no longer possible to immediately claim E θ = 0 => φ = constant on the boundary. We shall now show that, under suitable conditions, the last term -j ωAt is much smaller (in magnitude) than the first term - ∇tφ, and therefore we have Et ≈ -∇tφ and then φ ≈ constant by our argument above. An arm-waving argument is to say that F act 1 implies that the transverse potential A t can be neglected in a transmission line and therefore E t ≈ -∇tφ. But | At| << |Az| does not prove | ωAt| << |∇tφ | so we shall try to do better with a more substantial argument. First, we divide up the frequency domain (relative to some transmission line geometry) into a set of regimes. We state these for a round wire of radius a, but for a general conductor one can replace a with some typical transverse dimension of the conductor : δ > a/10 δ < a/10 δ < a/1000 low frequency strong skin effect extreme skin effect (3.7.3) Here δ ≡ 2/(ωμσ) is the skin depth of (2.1.8) or (2.2.20). Obviously the classification is arbitrary, we might have taken δ = a/5 as the strong skin effect boundary. We shall find that some facts which are approximately valid for the strong skin effect regime are almost exactly valid in the extreme skin effect regime. Here then is what we want to show: Fact 2: φ ≈ constant on a conductor surface in the strong or extreme skin effect regimes within the T r a n s m i s s i o n L i n e L i m i t . ( 3 . 7 . 4 ) See Comments below the proof regarding the significance of Fact 2. Our proof proceeds in a set of Steps (the Transmission Li ne Limit is defined in Step 4). As a guide, here is a little graphic showing how this proof works: Step 1. In the strong or extreme skin depth regime, A z ≈ (1/vd) φ . (3.7.5) As usual, the subscript "d" refers to a value in the dielectric between conductors, and here v d = 1/μdεd is the speed of light in the dielectric and also the phase velocity of a wave going down our transmission line. Similarly, βd = (ω /vd) is the wave's wavenumber in the dielectric. Using ∂z → -jβd as in (D.1.16) and our usual ∂ t → jω we find from (1.3.1) that Ez = -∂zφ - jωAz = jβdφ - jωAz = j(ω/vd)φ - jωAz = jω [ φ/vd - Az ] . βd = (ω /vd) => φ/vd - Az = Ez/(jω) ( 3 . 7 . 6 ) Chapter 3: Transmission Line Preliminaries 109 Inside a good conductor E z is small to begin with, and in the limit δ → 0 (ω→∞ ) the right side of (3.7.6) is small due to this fact and due to the 1/ ω factor. Therefore, A z ≈ φ/vd small or extreme skin effect (3.7.7) Sometimes a different argument is given to obtain (3.7 .7). In the King gauge we know from (1.5.5) that in the dielectric, div A = -j (β d2/ω)φ (1.5.5) or (∂ xAx+∂yAy) + ∂zAz = -j (βd2/ω)φ or (∂ xAx+∂yAy) - jβdAz = -j (βd2/ω)φ . (3.7.8) Without a proof, we extend the usual arm-waving argument that A t components can be neglected to say that transverse derivatives of At can also be neglected so ( ∂xAx+∂yAy) ≈ 0, and then we have - jβ dAz ≈ -j (βd2/ω)φ or A z ≈ (βd/ω)φ = φ/vd which replicates the conclusion (3.7.7) seemingly without the skin effect restriction. A more careful analysis must show that ( ∂xAx+∂yAy) can only be so neglected in the strong or extreme skin effect limits. Step 2. Claims that | ∇tφ| ≈ (1/D)|φ| where D is a characteristic transverse dimension of the transmission line. We might argue this on dimensi onal grounds alone, but consider |∂ xφ| ≈ Δφ Δx ≈ V D ≈ (1/D) |φ| . ( 3 . 7 . 9 ) This is a very crude use of the ≈ sign, there could be a factor of 10 or 1/10 on either side, but when combined with << in Step 4 below we still obtain a reasonable conclusion. Here V is the potential difference between the two transmission line conductors, and D is their "separation". Obviously | ∂xφ| is not the exact constant V/D at every point in space between the conductors, this is meant only as a ballpark estimate of the size of | ∂xφ| in some average sense. Step 3. Claims that | ωAx| << 2π |φ| (1/λ) where λ = traveling wave's wavelength. From Fact 1 we have, |A x| << |Az| . ( 3 . 7 . 1 0 ) Chapter 3: Transmission Line Preliminaries 110 With Step 1 (3.7.7) this says |Ax| << |φ| /vd = |φ| (βd/ω) or |ωA x| << (2π/λ) |φ| β d = 2π/λ (3.7.11) where λ is the wavelength of our transmission line wave. Step 4. Claims that | ωAx| << |∂xφ| In Chapter 4 we shall introduce the notion of the Tran smission Line Limit which is a requirement that on a transmission line, the wavelength λ must be much larger than any transverse dimension D of the line, λ > > D ( 3 . 7 . 1 2 ) or (1/λ) << (1/D) or 2π |φ| (1/λ) << 2π |φ| ( 1 / D ) . ( 3 . 7 . 1 3 ) Combining this with Step 3 (3.7.11) we find |ωA x| << (2π/λ) |φ| << 2π |φ| (1/D) or |ωA x| << 2π |φ| (1/D) . Bringing in the ballpark estimate Step 2 (3.7.9) that | ∂ xφ| ≈ (1/D) |φ| we then have |ωA x| << |∂xφ| where we just ignore the 2 π factor relative to our extreme << situation. Doing this also for y, we have |ωAt| << |∇tφ| . ( 3 . 7 . 1 4 ) Looking then at (3.7.2) one finds E t = -∇tφ - jωAt ≈ -∇tφ and this concludes our longw inded explanation of why φ ≈ constant on a transmission line conductor's cross section surface. We had to assume the Transmission Line Limit ( λ >> D) and we had to assume the strong or extreme skin effect regime to get φ ≈ constant. Chapter 3: Transmission Line Preliminaries 111 Comments: 1. Intuitive proof: We need high ω to get small δ. Currents in the thin δ surface sheath are in the z direction and there is "no room" for transverse currents in the sheath so Jt ≈ 0 and then At ≈ 0 so ωAt ≈ 0 so Et ≈ -∇tφ and then finally E θ = 0 => φ ≈ constant. 2. The fact that φ ≈ constant on each conductor surface embodies the electro-quasi-static transmission line theory. It will allow us to treat the transmissi on line as a "capacitor problem" in Chapter 5, as if we were doing electrostatics, even though we are at high ω and in the skin depth regime. 3. We know that φ = constant at ω = 0, but in order to prove that φ = constant at ω > 0 we had to make the extra assumptions stated above. We have not provided any proof that φ = constant for the "low frequency" range of (3.7.3), except for ω = 0. It seems likely that φ = constant is correct for very low frequencies close to ω = 0 and below some ω1, and probably φ ≈ constant is reasonable for the rest of the low frequency range (but we have not prove d this). Here then is the situation: very low ω low ω strong skin effect extreme skin effect 0 ≤ ω < ω 1 δ1 > δ > a/10 δ < a/10 δ < a/1000 φ = constant φ ≈ constant ? φ ≈ constant φ = constant (3.7.15) (d) B and A z on a conductor surface Fact 1 (3.7.1) says that |A x,y| << Az for a transmission line, and so we just set A x = Ay ≈ 0. In this case we find that B = curl A = x^ (∂yAz - ∂zAy) + y^ (∂zAx - ∂xAz) + z^ (∂xAy - ∂yAx) ≈ x^ (∂ yAz) + y^ (- ∂xAz) = Bt ( 3 . 7 . 1 6 ) which says B ≈ Bt is mainly in the transverse direction. In the extreme skin effect, we know that inside the conductor B decays to 0 quickly over distance δ (see Fig 2.9 for an isolated round wire). This is akin to the Meissner Effect where magnetic fields are ex cluded from the interior of a superconductor. In the extreme skin effect limit δ → 0, just below the thin surface current sheath we then have B n = 0 (since B = 0), where B n is the component of Bt normal to the surface. According to box (1.1.50) we know that B n is continuous through the boundary, so we must have B n = 0 just outside the surface as well. This is an application of div B = 0 ⇔ ∫S B • dS = 0 S is any closed surface (1.1.34) for a thin red Gaussian box shown here end-on on the left: Chapter 3: Transmission Line Preliminaries 112 On the left we imagine δ → 0 so the box can be made extremely thin so the left and right sides of the box then make no contribution to the flux. The front and back sides have no flux since B z ≈ 0 and because the sides are thin. Thus B n vanishes on the top face of the box since it vanishes on the bottom face. For finite δ on the right, this same thin box does not deliver this result. A more detailed argument would show that Bn = 0 to the extent that skin depth δ << r where r is the local radius of curvature of the surface. For a "perfect conductor" we have δ = 0 and B n = 0 exactly. We summarize our conclusions: Fact 3: (a) In general, the B field at a transmission line conductor surface has a negligible z component and so B ≈ Bt; (b) In the extreme skin effect regime, B ≈ Bt has no normal component B n at the conductor surface. This is approximately true in the strong skin effect regime. (3.7.17) Corollary: In the plane of a transmission line conductor cross section, and in the extreme skin effect regime, the magnetic field line pattern in the dielectric is such that just above the surface of each conductor there is a closed tangential B field line encl osing the conductor which is almost exactly parallel to the surface at every point. This fact is approxima tely true for the strong skin depth regime. (3.7.18) This is illustrated in the following figure where B field lines are shown in red: F i g 3 . 5 Fact 4: In a situation where A x,y can be neglected relative to A z we have seen that the B field lines are constrained to cross sectional planes. For any such planar set of B field lines, each B field line is an equipotential contour for A z. ( 3 . 7 . 1 9 ) Proof: Consider a small rectangular "math loop" into th e plane of paper (depth dz) as shown in the above Figure. The black segment shows this loop edge on. Since this loop is parallel to a B field line, the magnetic flux through the loop is zero. According to (1.1.39) we know that curl A = B ⇔ ∫{C A • ds = ∫S B • dS . (1.1.39) Chapter 3: Transmission Line Preliminaries 113 The line integral of A around our math loop must therefore vanish. But since A has only the component Az, the line integral has contributions only from the two sides of the loop (both of which are perpendicular to paper). Thus ∫{C A • ds = [ Az(1) - Az(2) ] dz = 0 so A z(1) = Az(2). By this argument, all points on the red B field line shown have the same value of A z and thus that red B field line is an equipotential contour for A z. But this applies to any of the red B field lines, so in general, each such B field line is an equipotential for A z. This is reminiscent of the fact that E field lines are equipotentials for φ in electrostatics. Fact 5: On each conductor boundary, A z ≈ constant in the extreme or strong skin effect regimes.(3.7.20) From (3.7.18) we know that in the extreme skin de pth regime, the innermost B field line almost exactly skirts the conductor perimeter. From (3.7.19) we know that any B field line is an equipotential contour. Thus, the cross section perimeter itself is very cl ose to an equipotential contour of the function A z(x,y,z). In the strong skin effect regime this constancy of A z on the boundary is only approximately true. Comment: In Fact 2 we argued that φ ≈ constant on a conductor perimeter in the extreme skin effect regime. We also argued in Step 1 that A z ≈ (1/vd)φ everywhere inside the conductor and therefore also at the conductor surface. Thus, Fact 2 that φ ≈ constant on the perimeter is c onsistent with Fact 5 that A z ≈ constant on the perimeter, and th ese two constants are related by A z ≈ (1/vd) φ. That is, A z( any point on perimeter) ≈ (1/vd) φ(any point on perimeter) //extreme δ (3.7.21) and for the strong δ regime, this is approximately true. A Counter Example and Comments o n the Low Frequency Regime We have argued above that in the strong/extreme skin effect regime, the perimeter of a transmission line conductor's cross section will align with a B fiel d line and will have a constant value of A z. This is in general not true for low frequencies. In particular, it is not true at ω = 0. As an example of this fact, we consider a pair of parallel round wires carrying current I and -I . Since the current density in the wires is uniform, it is an easy matter to compute B for each conductor and superpose to get the total B field due to both conductors. Here is a plot of the resulting magnetic field lines (details in Appendix O), Chapter 3: Transmission Line Preliminaries 114 F i g 3 . 6 a Notice that the red magnetic field lines, being loci of constant A z, do not align with the black conductor surfaces. One can conclude that the conductor surfaces are not surfaces of constant A z in this very low frequency example ( ω= 0). Now, having said this, we can make some very a pproximate low frequency remarks. In Chapter 5 we will arrive at following equations involving φ and Az and their transverse partners φt and Azt φ(x,y,z) = 1 4πε q(z) φt( x , y ) (5.1.1) Az(x,y,z) = μ 4π i(z) Azt( x , y ) . (5.2.1) [ ∇t2 + (βd2-k2)] φt(x,y) = 0 φt(C1) = K1 φt(C2) = K2 K1- K2 = K (5.3.10) [ ∇t2 + (βd2-k2)] Azt(x,y) = 0 A zt(C1) = W1 Azt(C2) = W2 W1- W2 = K . (5.3.11) These two boundary value problems assume the extreme skin effect regime so that φt and Azt are constants on both black circles. For the δ→ 0 skin depth limit, we expect then to have A zt(x,y) = φt(x,y). Using με = 1/vd2 and i(z) = q(z) v d, (4.11.19a) if one has A zt(x,y) = φt(x,y), then from the ratio of the first two equations above one also has A z(x,y,z) = (1/vd) φ(x,y,z) which is the Step 1 fact (3 .7.5) above (stated there for finite δ). The question then is this: to what extent is it true that A z ≈ (1/vd) φ in the low-frequency regime shown in (3.7.3), all the way down to ω = 0 ? Looking at Fig 3.6a, we can certainly find two red loci which are somewhat similar to our black conducto r boundaries, missing perhaps by 30%. These two red closed curves would then define a boundary value problem with a solution A zt that is roughly on the same scale as the solution A zt at high frequency. So our answer is this: Chapter 3: Transmission Line Preliminaries 115 Az ~ (1/vd) φ in the low frequency regime (3.7.5) low ω where ~ means both sides have the same general scale. Here is another version of Fig 3.6a in which the Az values of some of the red curves are shown, Fig 3.6b As an alternate to the above language, we could say that A z is ballpark constant on the right black circle, in that A z only varies from -0.4 to -1.25 (and not, say, from -.001 and -1000). How Fig 3.6b was made. Az(x,y) for one cylinder at DC is computed in Appendix B with result (B.7.7). We can superpose that A z with a similar A z for the other cylinder giving this result, Az(x,y) = μI 2π * { - lnr 1 θ(r1>a1) + [ (1-r 12/a12)/2 - lna1 ] θ(r1<a1) + lnr 2 θ(r2>a2) - [ (1-r 22/a22)/2 - lna2 ] θ(r2<a2) } . The drawing below shows the r i and ai with origin at the center of the left cylinder : Setting a 1 = a2 = 0.5 and b = 1.25, and ignoring the overall constant μI/2π, the plot was made using Maple's implicitplot call which is an x-y scanner producing a crude but acceptable plot: Chapter 3: Transmission Line Preliminaries 116 (e) Observations about the E and B field lines in a transmission line dielectric Fact 6 : In a cross sectional sketch of a tr ansmission line, the E field lines land on the conductors at right angles to the conductor surface. This is exactly true for the TEM mode, and applies to all points on the conductor surfaces. (3.7.22) Proof: This follows from Fact 1 (3.7.0) which says E θ = 0 at the conductor surface. Fact 7 : In a longitudinal sketch of a transmission line, the E field lines still land on the conductors at very close to right angles. (3.7.23) Proof: Although E θ = 0 at the conductor surface, E z is not zero, though it is very small. We know that E z exists inside the conductor to support J z = σEz , and we know by (1.1.41) that E z is continuous through the boundary, so the longitudinal E field landing angle will not quite be π /2. The deviation from π/2 is less than 10-4 radians according to (3.6.2), and the deviation is in the direction of current flow at each conductor. This causes a very slightly warping of the otherwise planar cross-sectional field line grid. Fact 8 : Apart from an overall scale factor, the cross-sectional field shape of a TEM wave on a transmission line is independent of position z along th e transmission line, and is independent of time t. The shape is also independent of ω . ( 3 . 7 . 2 4 ) Proof: As we shall see below, the TEM form of any field or current is F(x,y,z,t) = ej[ωt-kz+φF(ω)]F(x,y) where F(x,y) is real and all t and z dependence is in the exponential. We can take the physical field to be the real part as discussed in Section 1.6 so Fphysical (x,y,z,t) = cos[ ωt-kz+φF(ω)] F(x,y). Thus, the cross sectional shape of the field is determined by F(x,y) and is the same at all values of z apart from an overall scale factor cos[ ωt-kz+φF(ω)]. This scale factor varies between +1 and -1 as one moves down the line in z at some fixed t, or as one observes at some fixed z as time varies. Later we will see that this shape F(x,y) can be found by solving a certain 2D Helmholtz equa tion, and we find that the shape is determined entirely by the shape of the boundaries of the conductors. Different vector fields (e.g., J and E) might have different ω-dependent phases in this wave motion which we indicate by φF(ω) for F(x,y,z,t). Fact 9 : In a cross sectional sketch of a transmission line operating in the extreme skin effect regime, the E and B field lines are very nearly perpendicular at every point in the dielectric. In the strong skin effect regime, the fields are approximately perpendicular. (3.7.25) Proof: From Maxwell's curl E equation (1.1.2) in the ω domain we have curl E = - jω B . (1.1.2) Chapter 3: Transmission Line Preliminaries 117 Then B • E = (-jω )-1 curl E • E = (-j ω)-1 [ ( ∂xEy - ∂yEx)Ez + ( ∂yEz - ∂zEy)Ex + ( ∂zEx - ∂xEz)Ey ] . (3.7.26) In the extreme skin effect regime, for a given ω we think of conductivity σ being very large, and so the conductor's E z is very small. Since E z is continuous at the conductor boundary, E z is also very small in the dielectric. In contrast, due to the surface charge on the conductors, the transverse fields E x and Ey are very large in the dielectric. If we neglect E z and its derivatives in the above expression we find that B • E ≈ (-jω)-1 [ (- ∂zEy)Ex + ( ∂zEx)Ey ] ≈ (-jω)-1 [Ey2 ∂z(Ex/Ey) ] . ( 3 . 7 . 2 7 ) However, we argued in Fact 8 that the shape of fields does not vary with z. Thus, the ratio of two components like E x/Ey cannot vary with z, so ∂ z(Ex/Ey) = 0. Alternatively, we make the usual replacement ∂z → -jk (see Fact 8 proof) to get B • E ≈ (-jω)-1 [ (- ∂zEy)Ex + ( ∂zEx)Ey ] = (-jω)-1 [ (jkEy)Ex + ( -jkE x)Ey ] = (-j ω)-1 )(jk) [ (E y)Ex + ( -Ex)Ey ] = 0 . ( 3 . 7 . 2 8 ) We have already shown that at the conductor surfaces, E is normal to the surface and in the extreme skin effect regime B is nearly tangent to the surface, so we certainly have B • E ≈ 0 at the conductor surfaces. (f) Drawings of the fields We are now in a position to draw some sketches of fi elds on a transmission line. Let's start with the transverse or cross section picture: Fig 3.7 Fig 3.7: Cross section view Chapter 3: Transmission Line Preliminaries 118 Although this figure is drawn for two round conductors, its general features apply to any conductors. The figure is a snapshot at one instant in time. The • and ⊗ indicate current flow direction in the conductors. Positive charge exists on the surface of the left conduc tor, and is strongest on the face of that conductor which is closest to the other conductor. Negative surface charge lies on the right conductor. The electric fields are as shown and are strongest in the region between the conductors. The magnetic field directions derive from the right hand rule relative to the curre nt in each conductor. The lines of E and B always intersect at right angles as noted in (3.7.25). The magnitude of the E field is determined by th e potential difference between the conductors and the geometry. It is independent of frequency. Similarly, the magnitude of the B field is determined by the size of the current in either conductor a nd is also independent of frequency. Consider a 75 Ω transmission line that is properly terminated and is driven by a 7.5 volt amplitude sine wave. Regardless of frequency ω (but ω large enough so Z 0 = 75Ω, see (4.11.16)) , the magnitude of the current in this transmission line is 100 mA, and the magnitude of the potential difference is 7.5 volts. Of course both these quantities have sinusoidal time depe ndence. At some instant in time, the fields and currents are as in Fig 3.7. We have just argued then that not much happens in the transverse directions x and y as frequency sweeps up from strong skin effect to extreme skin eff ect. The rate at which the pattern oscillates back and forth increases, but the field pattern shape does not cha nge. This may seem contradictory. In general, one is used to ω affecting things due to equations like curl E = -jω B Maxwell curl E equation (1.1.2) The resolution is that all the spatial variation happens in the longitudinal direction. Here then is a top view of the same transmission line: Fig 3.8: Top view of transmission line Fig 3.8 Chapter 3: Transmission Line Preliminaries 119 The red E arrows are all of unit length and serve to mark the direction and density of electric field lines lying in the plane containing the center lines of the conductors. The blue B arrows are seen end-on and indicate the same for the magnetic field. On the left they come out of the plane of paper and on the right they go into it. Later we shall learn about the "transmission line limit" in which the wavelength λ of the wave propagating down a transmission line is assumed to be much larger than all transverse dimensions of the line. The reader should understand the above pi cture as being in that limit, but one would have to stretch the picture at least 10X horizontally to make it be reasona ble. At all places ExB points to the right, so we have a wave propagating to the right (+z). Now apply the Maxwell curl equations using the two loops shown. Loop 1 is positioned to pick up magnetic flux, so we use (1.1.36) which in the frequency domain says curl E = -jω B ⇔ ∫{ E • ds = -jω[∫S B • dS] (3.7.29) Notice the ω sitting on the right side. We argued in the last section that the amplitude of the B field does not change as ω changes. Thus, the right side of (3.7.29) is proportional to ω. As ω increases, the line integral of the E field around loop 1 must increase. Thus, the rate of change of E must increase in the z direction! In other words, as ω increases, the whole pattern of Fig 2 contracts in the z direction, which causes all z derivatives to increase, thus increasing ∫{E•ds for the same fixed loop 1. Remember that the strength of the E field is indicated in Fig 2 by the density of the red arrows, not by the length of the red arrows. A similar argument applies to loop 2. This loop appears end-on in Fig 3.8. It is set up to sense the electric field flux. The appropriate curl equation is (1.1.38) which says curl B = μ dεdjωE + μJc ⇔ ∫{ B • ds = μd ∫S [εdjωE + Jc] • dS ≈ jωμdεd ∫E•dS . (3.7.30) Since we are now in the dielectric, we have ignored the small leakage conduction current, and have kept the dominant displacement current. Again there is a factor of ω on the right side, arising from a time derivative. As ω increases, the line integral of the B field mu st increase. Thus, the B field must change faster in the z direction. As ω increases, the curl equation (3.7.30) is satisfied by having the entire pattern contract in the z dimension. If the frequency ω doubles, the wavelength λ goes to half. This of course is no surprise, since ω and λ are related by the speed of light ν d in the dielectric, λ = vd/f = 2πvd/ω . (3.7.31) The main point of the above discussion is to show how the Maxwell curl equations force the field pattern to contract in the z direction as ω increases. In the transverse direction, the field pattern shape stays constant. Chapter 3: Transmission Line Preliminaries 120 (g) More on the field and current structure Here we explore in more detail the general distributio n of fields and currents in a transmission line. The goal is to establish the phase relationships among the electromagnetic fields and various currents. Once this is done, it is possible to make an estimate of the ratio J r/Jz and that is done in the following section. Consider the following more elaborate version of Figure 3.8 : Tilted overhead view of a transmission line Fig 3.9 The picture is quite complicated and deserves clarifying comments: (1) Unlike in Fig 3.8, the E and B arrows indicate the E and B vectors, and are not just field direction and field line density indicators. (2) The E and B field vectors are shown along some line which lies in the plane of the center lines of the two conductors and which points in the z^ direction, as do those center lines. (3) The blue B field arrows lie in the blue plane which is meant to be perpendicular to the plane of the conductor center lines, which is the plane of paper. The red E field arrows are in the plane of paper. (4) Looking at E x B, we see that the wave is traveling to the right in the z^ direction. (5) The E field arrows point from positive charge to negativ e charge, so this is why the + and - signs are distributed as shown. Chapter 3: Transmission Line Preliminaries 121 (6) The conductors are fixed to the paper, everything else is moving to the right at velocity v d. This includes the E and B arrows and their curves, the charge dens ity and its curve n, and the two current curves drawn on the bottom conductor. (7) At point Q on plane z = z Q, since B is coming out of paper to the viewer, the longitudinal current J z in the lower conductor must be pointing to the right. This is why J z is shown positive at this point in the lower conductor, and this calibrates the position of the J z curve. Maximum J z occurs with maximum B. (8) There exists a displacement current Jdisp = ∂tD = εd ∂tE in the dielectric whose magnitude is shown as a red curve. For an observer sitting at fixed point P, since the wave is moving to the right, the value of ∂ tE is at its instantaneous maximum pos itive value. This is why the red J disp curve has a positive maximum at point P. (9) As discussed in Section 3.4, the displaceme nt current is "fed" by the radial current J r inside the lower conductor, so the J r curve also has its maximum positive value at point P. This J r current is busily radially pumping positive charge to the surface of the lower conductor at point P so that charge will be there when the wave has moved λ/4 to the right. Of course this radial J r is doing this charge pumping all around the lower conductor, but we only show it in the plane of paper. [ See Section D.9 (c) ] (10) We have glossed over the fact that the E and B fields track each other in magnitude. The Maxwell equation curl E = -jωB requires that E and B vanish at the same place (z = z P). Since E and B have the same wavelength, they must also have their maxima at the same place (z = z Q). This same correlation occurs in a normal plane wave. The maximum of E at z = z Q is associated with a maximum of the surface charge, while the maximum of B is associated with a maximum of J z. (h) Estimate of the ratio J r/Jz Having drawn and described this elaborate Fig 3.9, we now consider the inscribed green Gaussian box which contains no surface charge. We first assume cylindrical conductors so this box is a cylinder. At the instant in time for which Fig 3.9 is drawn, the tota l current flowing into the endcaps of the box is 2I, where I is the peak longitudinal current -- the magnit ude of the longitudinal sine wave. Therefore, the total J r integrated over the sides of the green cylinder must also be 2I. To obtain a ballpark estimate of the situation, we assume that the two round conductors are far apart compared to their radii, in which case J r is roughly symmetric around the conductor surface. Then the total radial current emitted by the curved surface of the green Gaussian cylinder is: radial current total = [ (2/ π)Jr ]* 2πa * (λ/2) = 2I Since Jr is a longitudinal sine wave, we have added a factor 2/ π to get its value averaged over the length of the Gaussian box. In a more general case, we can replace 2 πa with distance p which represents the active portion of the conductor perimeter, as illustrated in Fig 2.16. Then we have [ (2/π)Jr ]*p * (λ/2) = 2I => Jr = 2πI / (λp ) . ( 3 . 7 . 3 2 ) Chapter 3: Transmission Line Preliminaries 122 On the other hand, for a round conductor operating in the strong skin effect regime J z ≈ I/(pδ) ( 3 . 7 . 3 3 ) where p is the same active perimeter just mentioned. So Jr/Jz ≈ 2π (δ/λ) . ( 3 . 7 . 3 4 ) For δ we had δ ≡ 2/ωμσ . (2.2.20) From (3.7.31) we have λ = vd/f = 2πvd/ω where v is the wave phase velocity. Then (δ/λ) = 2 ωμσ ω 2πvd = 2ω μσ 1 2πvd = 4πf μσ 1 2πvd . (3.7.35) Setting v d ≈ c and μ = μ0 = 4π x 10-7 and σ = 5.81 x 107 (copper) and f = 109f(Ghz) we get (δ/λ) ≈ 4πf μσ 1 2πvd = 4π 109 f(GHz) 4π x 10-7 x 5.81 x 107 1 2π 1 3 x 108 = 1010 f(GHz) 58.1 1 2π 1 3 x 108 = f(GHz) 58.1 1 6π 10-3 = 7 x 10-6 f(GHz) and so Jr/Jz ≈ (2π) (δ/λ) ≈ 4.4 x 10-5 f(GHz) . (3.7.36) For f ≤ 10 GHz we then find Jr/Jz ≤ 1.4 x 10-4 . f ≤ 10 GHz strong skin effect regime (3.7.37) showing that the radial charge-pumping current density J r is much smaller than the longitudinal current density J z in the conductor sheath. What about the low-frequency situation with no skin-effect sheath? For simplicity, we assume now two round conductors of radius a which are wide ly spaced. No skin effect means roughly δ > a which means 2/ωμσ > a => ω < 2/(μσ a2) or ωa/2 < 1/(μσa) . (3.7.38) In this low frequency regime we must replace (3.7.33) by Jz ≈ I/(πa2) . ( 3 . 7 . 3 9 ) Chapter 3: Transmission Line Preliminaries 123 Since (3.7.32) is still valid, we find now that J z ≈ I/(πa2) Jr ≈ 2πI/(pλ) ≈ 2πI/(2πaλ) ≈ I/(aλ) so J r/Jz ≈ π(a/λ) ≈ (πa)(ω/2πvd) ≈ ωa/2vd = (ωa/2)(1/vd) . (3.7.40) Using (3.7.38) for ωa/2 we get Jr/Jz < 1/(μσavd) . ( 3 . 7 . 4 1 ) With μ = μ 0 = 4π x 10-7, σ = 5.81 x 107 (copper) and v d = c = 3 x 108 we find for a wire of radius 1 mm, Jr/Jz < 1 4π * 5.81 * 10-3 * 3 * 108 = 10-5 4π * 5.81 * 3 = 4.6 x 10-8 . low frequency (3.7.42) The conclusion is that in general J r << Jz under 10 GHz and finally we justify entries made in the tables of Sections 3.5 and 3.6. The basic fact is that the green cylinder in Fig 3.9 is long, so the surface area through which J r flows is much larger than the area through which J z flows. Comment : The explicit round wire field solution of Appendix D verifies that |J r/Jz| << 1. See (D.2.33) and Observation (3) following. Roughly the conclusion is that |J r/Jz| ≈ |βd/β'| << 1. 3.8 Review of Transmission Line Preliminaries A transm ission line normally has two conductors. Th e cross sectional shape of these conductors is assumed constant in the direction z along the transmission line. The transverse directions are x and y. A wave propagates down a transmission line in what is called the TEM mode. TEM means that the electric and magnetic fields of a wave traveling down the line are transverse, as in Figures 3.7-9. What this really means is that an electromagnetic wave goes straight down the conductors as guides with no surface reflections, unlike what happens in a wavegui de, see Appendix F. Apart from a small drag on the wave due to losses in the conductors, the wave proceeds with wavenumber β d and velocity νd as it would in an open medium. The conductors shape the E and B fields, so the wave is not a "plane wave". Nevertheless, at each point in the dielectric, E and B are perpendicular (strong skin effect regime) and E x B points down the transmission line. We now summarize a set of basic facts about this TEM mode, most of which were addressed in the previous Section. Fact 1 : The major current for the TEM mode is the longitudinal current J z. We just showed in the last section that J r << Jz. Moreover, J θ = σEθ vanishes at the surface from (3 .7.0) and is presumably either tiny or non-existent inside the conductor. (3.8.1) Chapter 3: Transmission Line Preliminaries 124 Fact 2 : There is no cutoff frequency one has to operate above. The TEM mode works all the way down to DC (although at low frequencies, the attenuation per wavelength becomes large). See Appendix F for why operation down to DC is not possible in a waveguide. (3.8.2) Comment : In the low frequency regime of (3.7.3) ther e is still a TEM wave going down the transmission line, but since we are not then in the strong or extreme skin depth limits, some of the facts of Section 3.7 do not apply. For example, looking at Fig 3.6, the conductors are no longer wrapped by tangent B field lines, and A z is no longer constant on the conductor perimeter, and E • B is no longer 0 at the surface. Corollary 2 : If one operates a transmission line below the cu toff of the lowest waveguide mode, the TEM mode is the only possible way of moving energy down the line. (3.8.3) Fact 3 : The simplest expression of the boundary conditions (at least for large ω ) are in terms of potentials , not fields, so the potential wave equations are used to solve problems. (3.8.4) Those boundary conditions are that φ and A z are constant on conductor cross sections at a given z, as stated below in Facts 6 and 7. Fact 4 : The transverse components of the vector potential A can be neglected, so A z is the only component of A we have to worry about. (3.8.5) Proof: This is addressed in (3.7.1) and Appendix M, but we give a brief summery here. Consider equation (1.5.9) where both conductors have the same μ, A(x,ω) = μ 4π ∫J(x',ω) e-jβdR R d V ' ( 3 . 8 . 6 ) Here, J represents the currents in the conductors and the volume integration is over both conductors in x,y and z, and R = | x-x'|. There is clearly going to be a strong A z component since the predominant conductor currents are in the longitudinal direction. According to Fact 1 above, transverse currents are very small, so the corresponding transverse components of A will also be very small and we shall completely neglect them. When we compute A in the above integral, we can still decompose A into A z, Ar and Aθ . These components are, however, with respect to some fi xed coordinate system located perhaps on some approximate center line between the two conduc tors. Thus, each potential of the pair A r and Aθ will feel the effect of both J r and Jθ , but these are both very small. Moreover, there is considerable cancellation which takes place as pieces of J r and Jθ are added up in the integration. We rely mainly on the fact that J r and Jθ are very small to conclude that A r and Aθ may be safely neglected. This is very different from what happens with A z. In the region of one conductor, the summation is additive for all nearby pieces of current J z in that conductor, assuming that the wavelength λ of longitudinal propagation is much larger than any transverse dimension. The only place A z is small is on a longitudinal line between the conductors where their contributions cancel. We conclude then that A r and Aθ can be neglected relative to A z. Chapter 3: Transmission Line Preliminaries 125 Fact 5: The potential φ(x) can be identified with the transv erse "voltmeter voltage" . (3.8.7) In the transverse direction (z = constant), and in the extreme/strong skin depth regime, we know from Fact 4 that Et = -∇tφ -jωAt ≈ -∇tφ because A t is very small. In the drawing below there is no difference then between the line integral of the elect ric field between the two black dots, and the potential difference φ1-φ2 between these same points. Since there is no B field perpendicular to the plane of paper, there is no time-varying magnetic flux through any loop containing the probe wires of our "planar" voltmeter, so there is no "EMF" induced in these l eads to confuse the meter reading, and the meter directly reads V = φ1-φ2. If in the drawing we move the right black dot attachment point to a point on the left conductor in some other z plane, the meter leads then enclose B field flux and -jωAz comes into play in Ez = -∂zφ - jωAz and it is then less clear what the voltmeter is reading. Fig 3.10 Fact 6 : The potential φ is constant over the surface of either conductor at a fixed z. (3.8.10) This was addressed in (3.7.4) where we had to add th e assumptions that we are in the strong or extreme skin effect regimes and we are operati ng in the transmission line limit. Although φ = constant at ω = 0, we concluded only that φ ≈ constant in the low frequency regime of (3.7.3). Fact 7 : The potential A z is constant over the surface of either conductor at a fixed z. (3.8.11) This was addressed in (3.7.20) and is only valid in the strong or extr eme skin effect regimes. At low frequencies Fact 7 is definitely not valid ( see Fig 3.6a). Chapter 4: Transmission Line Equations 126 Chapter 4: Transmission Line Equations In this Chapter we use the potential integral expres sions derived in Chapter 1 to derive the classical transmission line equations. We learn that most transmission line parameters are determined by a single geometric integral K. The approximations are clearly stated. 4.1 Computation of potential φ due to one conductor of a trans mission line Our starting point is the potential φ expression given in box (1.5.23) for the potential at some arbitrary point x in the dielectric due to conductor C 1 of a transmission line, φ1(x,ω) = 1 4πξd ∫ C1 ρ1(x',y',z',ω) e-jβdR R dx'dy'dz' . R = | x - x'| (4.1.1) Here the point x' = (x',y',z') runs over the surface of C 1 and R is the distance between the observation point x in the dielectric and the integration point x'. Parameters β and ξ are for the dielectric. Comments on ρ 1: 1. ρ 1 is the volume charge density associated with "surface charge" n 1 according to ρ1dV' = n1dS' . 2. ρ1 is a distribution. For example, for a round wire of radius a we expect ρ1 to be proportional to δ(r'-a) where r' = x'2+y'2 . Perhaps ρ1 = f(θ')δ(r'-a) where (r', θ',z') are cylindrical coordinates with axis at the round wire center. 3. Recall from Section 1.5 (c) and (1.5.17) the fact th at there are two distinct areal charge distributions called n c and ns which are related by n c = (ξd/εd)ns. Here n s is the actual surface charge distribution, whereas n c is an adjusted charge density which is directly associ ated with the current I in the conductor and which accounts for possible leakage in the dielectric. Our n 1 and ρ1 are associated with this n c adjusted charge distribution, not with n s. That is why the external factor in (4.1.1) is 1/4 πξ•d instead of 1/4πεd. Consider now this charge density ρ1(x). Following a standard methodology, we make the assumption that its functional form may be factored in the following manner, ρ 1(x,y,z) = α1(x,y) q1( z ) . ( 4 . 1 . 2 ) C/m3 1/m2 C/m The dimensions of the functions in this factori zation are as indicated, so the charge goes with q 1. Moreover, without any loss of generality we select the relative scale of the two factors such that the integral of α1(x,y) over a slice of conductor C 1 at any z is unity, ∫C1 dx dy α1( x , y ) = 1 . ( 4 . 1 . 3 ) Chapter 4: Transmission Line Equations 127 Therefore, we can interpret q 1(z) as the total charge per unit length on C 1 at location z : ∫C1 dx dy ρ1(x,y,z) = q 1(z) ∫C1 dx dy α1(x,y) = q 1(z) • 1 = q 1(z) . Assume that q 2(z) is the charge on the other conductor C 2 of a two-conductor transmission line. If q 1(z) + q2(z) ≠ 0, then we have a net charge per unit length and the transmission line is acting as a radiating antenna as well as a transmission line. From now on, we ignore this superposed radiation problem and assume that at each value of z, the net charge on both conductors is 0 -- the line is "balanced". Thus, q 2(z) = - q1(z) ≡ -q(z) . (4.1.4) To simplify notation, we now dispense with the subscript and denote q 1(z) = q(z). However, we maintain the subscript on α 1(x,y) to emphasize that the two conductors can have completely different cross sectional shapes. The shape of the transverse distribution of charge on C 1 is determined by α1(x,y), but the total charge is q(z) per unit length. How can we justify assumption (4.1.2)? This is "separation of variables". The idea is that we assume it without any justification, and then we try to find a solution to our problem which is consistent with the assumption. All we really want is to find a solution to our basic differential e quations with their boundary conditions, and any assumptions we ma ke can be justified in the end once we have found a solution. On the other hand, if an assumption like (4.1.2) does not l ead to a solution, then it must have been a bad assumption. We have seen earlier how the expected EM field pattern on a transmission line has a constant transverse "shape" and this certainly motivates the assumption (4.1.2). Now insert (4.1.2) into (4.1.1) to get, φ1(x,y,z) = 1 4πξd ∫-∞ ∞ dz' q(z') ∫C1 dx' dy' α1(x',y') e-jβdR R (4.1.5) R2 = (x-x')2 + (y-y')2 + (z-z')2 . 4.2 Computation of potential φ due to both conductors of a tran smission line Let us now write the potential at an arbitrary point x in the dielectric due to both conductors C 1 and C2 : φ12(x) = φ1(x) + φ2(x) = 1 4πξ ∫-∞ ∞ dz' q(z'){ ∫C1 dx1' dy1' α1(x1',y1') e-jβdR R – ∫C2 dx2' dy2' α2(x2',y2') e-jβdR R } R 12 = (x-x1')2 + (y-y1')2 + (z-z')2 = s12 + (z-z')2 s 12 = (x-x 1')2 + (y-y1')2 (4.2.1) R 22 = (x-x2')2 + (y-y2')2 + (z-z')2 = s22 + (z-z')2 s22 = (x-x 2')2 + (y-y2') . The minus sign between the terms is due to (4.1.4 ). Each conductor has its own arbitrary transverse charge distribution αi. The transverse integration variables on C 1 are dx1' dy1', while those on C 2 are Chapter 4: Transmission Line Equations 128 instead dx 2' dy2'. In the last two lines we intr oduce certain transverse distances s 1 and s2 as shown. The same dz' integration variable is used for both conductors. The following drawing shows an arbitrary dielectric point x = (x,y,z) located in the z = z plane. The point x1' = (x1',y1',z') lies on C 1 at some point of the C 1 integration and similarly for x2' = (x2',y2',z'). The full distances R 1 and R2 and the transverse distances s 1 and s2 are shown. Fig 4.1 One can imagine an expression similar to (4.2.1) fo r a transmission line consis ting of N conductors where Σi=1Nqi(z) = 0, but we shall restrict our interest to N = 2. 4.3 The Transmission Line Limit Consider again the potential at x due to both conductors shown in (4.2.1), φ12(x) = 1 4πξ ∫-∞ ∞ dz' q(z'){ ∫C1 dx1' dy1' α1(x1',y1') e-jβdR R – ∫C2 dx2' dy2' α2(x2',y2') e-jβdR R } ( 4 . 3 . 1 ) As the red-dashed z = z' plane shown in Fig 4.1 is pus hed back far from the z = z plane, the vectors which are labeled by distances R 1 and R2 become more aligned, and both R 1 and R2 become larger. During the transverse integrations over x1' and x2', these R i vectors then don't vary much. One could then replace the transverse charge density α1(x1',y1') with a point charge at the "center of the conductor" and not make much difference in the R1 vector and its length R 1. In this situation, the {...} integrand of the above integral has this form { e -jβdR1 R1 – e-jβdR2 R2 } . // when |z-z'| is large (4.3.2) If we then expand the exponentials show ing the first few terms, this becomes Chapter 4: Transmission Line Equations 129 { 1-jβdR1+(jβd)2R12/2 R1 – 1-jβdR2+(jβd)2R22/2 R2 } = {[ 1 R1 - 1 R2 ] + [-jβd + jβd ] + (jβd)2/2 [R1-R2] + ...} = { [ 1 R1 - 1 R2 ] - (βd2/2) [R1-R2] + order( βd3) } (4.3.3) Since R1 ≈ R2 for large |z-z'| as just discussed, both the leading term and the βd2 term are small in an absolute sense as long as βd2 is not huge. When |z-z'| is large, both R 1 and R2 are large and thus both 1/R 1 and 1/R2 are small, and [ 1/R 1 - 1/R2] is smaller still due to cancellation between the terms. So our first point is that, in the dz' integration, the main contribution to φ12(x) comes from regions of z' for which |z-z'| is small. Given then that the dz' integration in (4.3.1) is dominat ed by that part for which |z-z'| is small, we can see that for this controlling integration region the size of distances R 1 and R2 will be on the order of the transverse dimension of the transmission line, assuming that we select the point x somewhere between the two conductors. If we vaguely define the transmission line's transverse extent as distance D, then suppose we make the following assumption concerning βd : βdD << 1 "small βd" . (4.3.4) In this case, we can replace e -jβdR1 = 1 and e-jβdR2 = 1 in the integration without significantly changing the result. Then as shown in (4.3.3) there will be a correction term that is order βd2 which we shall neglect, as well as higher terms of order βdn with n> 2. Notice that the linear βd term vanished exactly in our large |z-z'| analysis. This linear term also vanishes in the full analysis since the α i transverse charge functions are normalized to unity: { ∫C1 dx1' dy1' α1(x1',y1') -jβdR1 R1 – ∫C2 dx2' dy2' α2(x2',y2') -jβdR2 R2 } = (-j βd) { ∫C1 dx1' dy1' α1(x1',y1') - ∫C1 dx1' dy1' α1(x1',y1') ) = (-jβd) {1 - 1} - 0. (4.3.5) Thus, by setting βd = 0 in (4.3.1) we are ignoring corrections on the order of βd2 and higher, and if βd is small, these corrections are very small. The Helmholtz parameter β d for the dielectric is 2 π/λ where λ is the wavelength of a wave passing down the transmission line. Thus, our "small βd" assumption stated above can also be written λ > > D ( 4 . 3 . 6 ) which says the wavelength is much longer than the si ze of the transmission line transverse dimensions. This assumption is called the Transmission Line Limit . If we operate within this limit, then (4.3.1) may be approximated as φ 12(x) = 1 4πξd ∫-∞ ∞ dz' q(z'){ ∫C1 dx1' dy1' α1(x1',y1') 1 R1 – ∫C2 dx2' dy2' α2(x2',y2') 1 R2 } (4.3.7) Chapter 4: Transmission Line Equations 130 We shall now use the small β d assumption one more time. We assume that the linear charge density q(z') has the characteristics of a wave traveling dow n the transmission line (see also Chapter 5), q(z) = q(0) e-jβdz // q(z,t) = q(0,0) ej(ωt-βdz) (4.3.8) so that q'(z) = -j β d q(z) q"(z) = (-j βd)2q(z) and so on. We can then write a Taylor expansion for charge density q(z') which appears in our integration, q(z') = q(z) + (z'-z) q'(z) + (1/2) (z'-z)2 q"(z) + ... = q(z) + (-j βd) q(z) (z'-z) + (1/2) (-jβ d)2 q(z) (z'-z)2 + ... = q(z) [ 1 + (-j βd) (z'-z) + (1/2) (-j βd)2(z'-z)2 + ... ] . (4.3.9) Since both R 1 and R2 are even functions of the quantity (z'- z), and since there is no other (z'-z) dependence in the (4.3.7) integrand, the (-j βd) term in (4.3.9) contributes nothing (this is also true more generally for (4.3.1)). Thus, if we assume small βd, we can approximate q(z') ≈ q(z) where we are then ignoring a βd2 size term. Once again, if βd is small, βd2 is very small so our error in replacing q(z') by q(z) is very small. We are only interested in th e contributing region where |z-z'| is on the order of transverse dimension D, so the same βdD << 1 is being assumed as earlier. We arrive then at our final result for the potential at a point x between the conductors, φ12(x) = 1 4πξd q(z) ∫-∞ ∞ dz'{ ∫C1 dx1' dy1' α1(x1',y1') 1 R1 – ∫C2 dx2' dy2' α2(x2',y2') 1 R2 } (4.3.10) where we have thrown out terms of order β2 and higher. In this transmission line limit approximation, our Helmholtz integral (4.3.1) has been reduced to essentially an electrostatics Coulomb integral where we just sum over the contribution of each piece of charge to the total potentia l. As noted earlier, q(z) has the normalization of n c and not n s as discussed in Section 1.5 (c) which explains why the leading factor is 1 4πξd and not 1 4πεd . This allows for the dielectric to have some conductance. It should be noted that the integral of (4.3.10) co nverges due to the subtraction of the two terms which in turn results from the two conductors having oppos ite longitudinal charge densities. The individual terms in (4.3.10) do not converge and are in fact each logarithmically divergent in the sense ∫-∞ ∞ dz' (1/z') = ∞ . Chapter 4: Transmission Line Equations 131 4.4 General Calculation of V(z) We now intr oduce two new points x1 and x2. The point x1 lies on C 1 in the z = z plane, while x 2 lies on C2 in this same plane. We then evaluate φ12(x) at x = x1 and subtract from that φ12(x) at x = x2 and in this way we obtain the potential di fference between the surfaces of the two conductors at z = z. Recall, Fact 2: φ ≈ constant on a conductor surface in the strong or extreme skin effect regimes within the T r a n s m i s s i o n L i n e L i m i t . (3.7.4) Thus, assuming the small δ regime and treating φ ≈ constant as an equality, the potential difference will be independent of the locations of x 2 and x1 as long as they are on their respective surfaces and both have z = z. For this reason, the potential difference is a f unction only of z. Thus we write, using two copies of (4.3.10), V(z) ≡ φ 12(x1) - φ12(x2) = 1 4πξd q(z) ∫-∞ ∞ dz' { ∫C1 dx1' dy1' α1(x1',y1') 1 R11 – ∫C2 dx2' dy2' α2(x2',y2') 1 R12 } – 1 4πξd q(z) ∫-∞ ∞ dz' { ∫C1 dx1' dy1' α1(x1',y1') 1 R21 – ∫C2 dx2' dy2' α2(x2',y2') 1 R22 } (4.4.1) where R112 = (x1-x1')2 + (y1-y1')2 + (z-z')2 = s112 + (z-z')2 s 112 = (x1-x1')2 + (y1-y1')2 R122 = (x1-x2')2 + (y1-y2')2 + (z-z')2 = s122 + (z-z')2 s 122 = (x1-x2')2 + (y1-y2')2 R222 = (x2-x2')2 + (y2-y2')2 + (z-z')2 = s222 + (z-z')2 s 222 = (x2-x2')2 + (y2-y2')2 R212 = (x2-x1')2 + (y2-y1')2 + (z-z')2 = s212 + (z-z')2 s 212 = (x2-x1')2 + (y2-y1')2 . (4.4.2) The vector R 12 points from our new point x1 to an integration point x2' on C2. Here is a drawing of our new and more complicated situation: Fig 4.2 Chapter 4: Transmission Line Equations 132 We next rearrange the four terms in (4.4.1) to get V ( z ) ( 4 . 4 . 3 ) = q(z) 1 4πξ d ∫-∞ ∞ dz' { ∫C1 dx1' dy1' α1(x1',y1')( 1 R11 - 1 R21 ) - ∫C2 dx2' dy2' α2(x2',y2') ( 1 R12 - 1 R22 ) } . It is now possible to carry out the dz' integrations. The integral of interest is the following, ∫-∞ ∞ dx { 1 a2+ x2 - 1 b2+ x2 } = ln(b2/a2) . (4.4.4) Since this is quite important, we confirm with Maple, The separate integrals here are logarithmically divergent, but the combination converges. Thus, ∫-∞ ∞ dz' ( 1 R11 - 1 R21 ) = ∫-∞ ∞ dz' (1 s112 + (z-z')2 - 1 s212 + (z-z')2 ) = ln(s 212/s112) and ∫-∞ ∞ dz' ( 1 R12 - 1 R22 ) = ln(s 222/s122) ( 4 . 4 . 5 ) so that V(z) = q(z) 1 4πξd { ∫C1 dx1' dy1' α1(x1',y1') ln(s212/s112) - ∫C2 dx2' dy2' α2(x2',y2') ln(s222/s122) } s212 = (x2-x1')2 + (y2-y1')2 s 222 = (x2-x2')2 + (y2-y2')2 (4.4.6) s112 = (x1-x1')2 + (y1-y1')2 s 122 = (x1-x2')2 + (y1-y2')2 . The four transverse distances are shown in this figure, Chapter 4: Transmission Line Equations 133 Fig 4.3 Equation (4.4.6) expresses the potential between the two transmission line conductors at some plane z in terms of the charge distributions on the conductors αi. In general, these charge distributions are not known, so one cannot regard (4.4.6) as a general purpo se silver bullet to solve transmission line problems. On the other hand, as we shall see, equation (4.4.6) is one of a group of equations which will allow us to express several different transmission line parameters in terms the same integral, and one then obtains a relation between these parameters. For example, in analogy to what we did with a pa rallel plate capacitor in (1.5.19), we may define the complex capacitance C' per unit length of our transmission line using (4.4.6) as follows: 1 C' = V(z) q(z) = 1 4πξd { ∫C1 dx1' dy1' α1(x1',y1') ln(s212/s112) - ∫C2 dx2' dy2' α2(x2',y2') ln(s222/s122) } = 1 4πξd K and V(z) = q(z) 1 4πξd K (4.4.7) where K is a dimensionless real number obtained from a geometric integral of the normalized transverse charge distributions αi (recall that αi is has dimensions 1/m2 in (4.1.2)), K ≡ ∫C1 dx1' dy1' α1(x1',y1') ln(s212/s112) - ∫C2 dx2' dy2' α2(x2',y2') ln(s222/s122) . (4.4.8) Recall from (1.5.20) that (C', C and G ar e discussed further in Section 4.11 below) C' = C + 1/(j ωR) = C + G/(j ω) ( 4 . 4 . 9 ) where conductance (per unit length) G is associated with the imaginary part of C'. We then have 4πξ d/K = C' = C + G/j ω or 4π(ε d+σd/jω)/K = C + G/j ω // (1.5.1a) for ξd Chapter 4: Transmission Line Equations 134 so that C = 4 πε d/K capacitance per unit length of the transmission line G = 4 πσd/K conductance per unit length of the transmission line so C/G = εd/σd . ( 4 . 4 . 1 0 ) Here G = 1/R' is the conductance across the dielectric between a unit length of the two conductors. This is unrelated to the longitudinal resistance R of the conductors themselves, though that parameter will arise later on in the form of surface impedance Z s. We only have G ≠ 0 if the dielectric has some conductance σd ≠ 0 ( or σ eff ≠ 0 as in (3.3.4) ). Note from above and (1.2.8) that dim(C) = dim( ε) = farad/m and dim(G) = dim( σ) = mho/m . We now quote several results that will be derived later in Section 4.11. First we show below in (4.11.29) with (4.11.30) that the external inductance per unit length of our transmission line is also related to this same constant K, L e = (μd/4π) K . ( 4 . 4 . 1 1 ) Second, we show in (4.11.16) that the characteristic impedance of the transmission line is given by Z0 ≡ V(z) i(z) = R + jωL G + jωC . (4.4.12) At sufficiently large ω we can neglect the R and G terms to get this real value, Z0 = L C . // large ω ( 4 . 4 . 1 3 ) Third, we show in (4.11.36) that, for large ω, L → Le so Z0 = Le C = (μd/4π)K 4πεd/K = (1/4π) K μd/εd = (1/4 π) K Zm (4.4.14) where Z m = μd/εd is the "intrinsic impedance of the dielectric medium" having μd and εd. Recall that for free space we had in (1.1.29) Zfs = μ0/ε0 = 376.73032 ohms => Z fs /4π = 29.97948 ≈ 30 Ω . (4.4.15) Typically one has μd = μ0 so then (note that εrel and K are dimensionless), Chapter 4: Transmission Line Equations 135 Z0 = (1/4π) K μ0/εd = (1/4π) K μ0/ε0 ε0/εd = (1/4π) K Zfs ε0/εd = K (Z fs /4π) / εd/ε0 or Z0 ≈ (K /εrel ) 30Ω . εrel ≡ εd/ε0 . (4.4.16) We then summarize the parameters of a transmission line in terms of dimensionless real integral K : C = 4 πε d/K capacitance per unit length (4.4.17) G = 4 πσ d/K transverse conductance per unit length L e = (μd/4π) K external inductance per unit length Z0 ≈ (K /εrel ) 30Ω characteristic impedance μd = μ0, εrel ≡ εd/ε0 R = Re(Z s1+Zs2) resistance of conductors, see (4.11.34) L = L e + (1/ω) Im(Zs1+ Zs2) total inductance, see (4.11.34) Zsi = surface impedance of conductor i, see (2.4.1) and (4.11.10) The last three items are not determined by integral K and we just mention them for completeness's sake. All these equations will be more fully developed in S ection 4.11 below, but we jump ahead a bit in order to display two important examples. Intrinsic Impedance Notice that the intrinsic impedance of a dielectric medium Z m = μd/εd is different from the characteristic impedance of a transmission line Z 0, although the two numbers are in the same ballpark. For large ω, we show above that they are related by the equation Z 0 = (K/4π) Zm. Both have dimensions of ohms (not ohms/m). One can define a different intrinsic impedance Z m = μd/ξd [ recall (1.5.1a) that ξd ≡ εd - jσd/ω ] and corresponding characteristic impedance Z0 which have the relationship Z0 = L/C' = (1/4 π) K Zm with C' as shown in (4.4.9) above. Be lden sometimes refers to to Zm as η . We shall have no use for the quantities Z0 and Zm since we handle conducting dielectrics without involving these quantities. 4.5 Example: Transmission line with wi dely-spaced round wires of unequal diameters Consider a transmission line made from two round wires of radii a 1 and a2 and center line spacing b. In the case that b >> a 1 and a2, the charge distribution on each round wire is symmetric about the wire and in this situation ( a rare one admittedly) we know the two transverse charge distributions: α1(x,y) = α1(r,θ) = δ(r - a1)/(2πa1) α2(x,y) = α2(r,θ) = δ(r - a2)/(2πa2) . (4.5.1) Chapter 4: Transmission Line Equations 136 The 1/(2πa1) factor is required so that the integral of α1 is unity as required by (4.1.3), ∫C1 dx dy α1(x,y) = ∫0 2π dθ ∫0 ∞ rdr α1(r,θ) = ∫0 2π dθ ∫0 ∞ rdr δ(r - a1)/(2πa1) = ∫0 2π dθ a1/(2πa1) = 2π a1/(2πa1) = 1 . (4.5.2) Note : The reason the αi are symmetric is that the two conducto rs are so far apart that each one is essentially "in isolation" and so the charge assumes an axially symmetric distribution. An analogy would be that for two point charges far apart, the E field close to either point charge is spherically symmetric because close to one charge the field of the other can be neglected. Our task is then to compute the integral K shown in (4.4.8), K = ∫0 2π dθ1 ∫0 a1 r1dr1 [δ(r1 - a1)/(2πa1)] ln(s212/s112) - ∫0 2π dθ2 ∫0 a2 r2dr2 [δ(r2 - a2)/(2πa2)] ln(s222/s122) . = ∫0 2π dθ1 1/(2π) ln(s212/s112) - ∫0 2π dθ2 1/(2π) ln(s222/s122) = 1/(2 π) { ∫0 2π dθ1 [ln(s212) - ln(s112)] - ∫0 2π dθ2 [ln(s222) - ln(s122)] } (4.5.3) where we then have four integrals to evaluate. We shall choose our V(z) potential-determining reference points x 1 and x2 as shown in this drawing, Fig 4.4 Chapter 4: Transmission Line Equations 137 The four s ij distances can be read off from the drawing using the law of cosines, s212 = a12 + (b-a1)2 - 2 a1(b-a1) cos(θ1) s 112 = a12 + a12 - 2 a1 a1 cos(θ1) = 2a12(1 - cos(θ 1)) s 222 = a22 + a22 - 2 a2 a2 cos(π -θ2) = 2a22(1 + cos(θ2)) s 122 = a22 + (b-a2)2 + 2 a2(b-a2) cos(θ2) . (4.5.4) We then invoke the following integral from p 531 of GR7, which we rewrite as ∫0 2π dθ ln (A ± Bcos θ) = 2π ln[(1/2)(A + A2-B2 )] . (4.5.5) The four integrals are then easily evaluated: ∫0 2π dθ1 ln(s212) = ∫0 2π dθ1ln([a12 + (b-a1)2 - 2 a1(b-a1) cos(θ1)] A = a 12 + (b-a1)2 B = 2 a 1(b-a1) A2-B2 = [a12 + (b-a1)2]2 - 4 a12(b-a1)2 = [a12 - (b-a1)2]2 => A2-B2 = (b-a1)2- a12 > 0 b >> a 1 => ∫0 2π dθ1 ln(s212) = 2π ln[(1/2)( a 12 + (b-a1)2 + (b-a1)2 - a12 ) = 2π ln[(b-a1)2] . The fourth integral is the same with 1 ↔ 2, and the different sign of the second term in s 122 makes no difference, ∫0 2π dθ2 ln(s122) = 2π ln[(b-a2)2] . The second integral is ∫0 2π dθ1 ln(s112) = ∫0 2π dθ1ln([2a12(1 - cos(θ 1))] A = B = 2a 12 , A2-B2 = 0 = 2 π ln[(1/2) 2a 12] = 2π ln(a12) . Chapter 4: Transmission Line Equations 138 The third integral is similar giving ∫0 2π dθ2 ln(s222) = 2πln(a22) . To summarize: ∫0 2π dθ1 ln(s212) = 2π ln[(b-a1)2] ∫0 2π dθ1 ln(s112) = 2π ln(a12) ∫0 2π dθ2 ln(s222) = 2πln(a22) ∫0 2π dθ2 ln(s122) = 2π ln[(b-a2)2] . (4.5.6) Then from (4.5.3) we find K = 1/(2 π) { ∫0 2π dθ1 [ln(s212) - ln(s112)] - ∫0 2π dθ2 [ln(s222) - ln(s122)] } = l n [ ( b - a 1)2] - ln(a12) - ln(a22) + ln[(b-a 2)2] = ln [(b-a1)2(b-a2)2 a12a22 ] = 2 l n [(b-a1)(b-a2) a1a2 ] = 2 ln [b2 a1a2 ] // since we assumed at the start that b >> a 1, a2 = 4 l n ( b / a1a2 ) . ( 4 . 5 . 7 ) Therefore the transmission line parameters from (4.4.17) are, K = ln(b/ a1a2 ) / / b > > a 1, a2 C = 4 πεd/K = πεd / ln(b/ a1a2 ) G = 4 πσd/K = πσd / ln(b/ a1a2 ) Le = (μd/4π) K = (μ d/π) ln(b/ a1a2 ) Z0 = (K /εrel ) 30Ω = (1/ εrel ) ln(b/ a1a2 ) 120Ω . (4.5.8) Sometimes these formulas are written in terms of wire diameters d i = 2ai in which case K = 4 ln[b/ a1a2 ] = 4 ln[2b/ d1d2 ] = 2 ln[4b2/d1d2] . (4.5.9) Since we are assuming b >> d 1,d2 we know that x ≡ 2b2/d1d2 >> 1. Therefore ch-1x = ln[x + x2-1 ] ≈ ln(2x) // an identity Siegel 8.56, then an approximation (4.5.10) so ch-1(2b2/d1d2) ≈ ln(4b2/d1d2) . Chapter 4: Transmission Line Equations 139 Then we can write K as K = 2 ln[4b 2/d1d2] = 2 ch-1(2b2/d1d2) ( 4 . 5 . 1 1 ) and so Z0 = (K /εrel ) 30Ω = (1/ εrel ) ch-1(2b2/d1d2) 60Ω . (4.5.12) It is not easy to find expressions for C,G and L e for the unequal radii geometry, but Z 0 does appear for example in Reference RDE page 29-23 where we find: Fig 4.5 with D = our b. For D >> d 1,d1 this shows N = 2D2/(d1d2), and this then agrees with (4.5.12). This quoted result is in fact correct (with the two extra te rms shown in N) even when D is not large. We shall derive this full result in Chapter 6, equation (6.3.12). In the special case that a 1 = a2 ≡ a we get, K = 4 ln(b/a) C = 4 πεd/K = πεd/ ln(b/a) G = 4 πσd/K = πσd/ ln(b/a) Le = (μd/4π) K = ( μd/π) ln(b/a) Z0 = (K /εrel ) 30Ω = (1/ εrel ) ln(b/a) 120 Ω = (1/ εrel ) ln(2b/d) 120 Ω d = 2a . (4.5.13) The first three results agree with King TLT p17 (30b), The expression for Z 0 agrees with the RDE source quoted above, Chapter 4: Transmission Line Equations 140 Fig 4.6 where again D = b and εrel= 1. Power Transmission Lines (also Telephone and Telegraph) Ignorin g proximity effects of the ground and possibl e ground wires, one can consider a single phase power transmission line as fitting into this example. The first interesting number is skin depth. For aluminum at f = 60 Hz we find σ aluminum = 3.7 x 107 mho/m // recall σcopper ≈ 5.8 x 107 (annealed) μ0 = 4π x 10-7 henry/m δ ≡ 2/(ωμσ) ≈ 2/(2πfμ0σ) = 1/(πfμ0σ) So δ ≈ 1 cm. Thus, skin effect could be significant for a very large diameter wire. Typically the individual strands of a 1500 amp cable are 1/6" in diameter or 0.2 cm in radius, so there is some slight non-uniformity in the current distribution. If the strands are not insulated one should think of this more in terms of the total cable diameter includi ng all strand layers which might be 1". Usually the requirement of low power loss requires that R be relatively small compared to ωL. The 1500A cable just noted has R = .02 Ω per thousand feet. Similarly, the conductance G (mostly from insulator leakage) is very small compared to ωC. Thus, (4.4.12) leads to (4.4.16) stating Z 0 ≈ K 30Ω . If the full cable is 1" in diameter and the two lines ar e spaced 1 m apart, we can compute K from (4.5.13), K = 4 ln(b/a) = 4 ln( 1m/0.5") = 4 ln(39.37*2) = 17.5 so then from (4.4.16), Z 0 ≈ K 30Ω = 17.5 * 30 Ω = 524Ω Notice that halving radius a (or doubling b) increases K by ln2 = 0.7 which is only 4% of 17.5, so Z 0 is fairly insensitive to the line geometry. Rajput (p 554) claims power lines typically range from 400 to 600 Ω. See southwire.com for data on transmission line cables. A twin-line telegraph or telephone cable falls into this same impedance class, with 600 ohms being the traditional Z 0 number. A single telegraph wire over the ground plane has a similar Z 0. For a = 0.5 cm and height 4 m, K = 2 ln(2h/a) from (6.3.21) below, so K = 2 ln(8/[.5x10-2]) = 2 ln(1600) = 14.8 giving Z0 = 440 Ω. Chapter 4: Transmission Line Equations 141 4.6 Example: A coaxial cable A coaxial cab le is the other transmission line where we know the surface charge distribution is that given by (4.5.1). The analysis of the pr evious section resulting in (4.5.3) is then unchanged, and we find that K is still given by (4.5.3), K = 1/(2 π) { ∫0 2π dθ1 [ln(s212) - ln(s112)] - ∫0 2π dθ2 [ln(s222) - ln(s122)] } . (4.5.3) What is different is that we have a different picture describing the various s ij distances. The new picture is this, where the cross section circles have radii a 2 > a1 : Fig 4.7 As we did in the previous section, we "read off" the s ij expressions using the law of cosines: s 212 = a12 + a22 - 2 a1a2cos(θ1) s 112 = a12 + a12 - 2 a1 a1 cos(θ1) = 2a12(1 - cos(θ 1)) s 222 = a22 + a22 - 2 a2 a2 cos(θ2) = 2a22(1 - cos(θ 2)) s 122 = a22 + a12 - 2 a1a2 cos(θ2) . ( 4 . 6 . 1 ) Recalling, ∫0 2π dθ ln (A ± Bcos θ) = 2π ln[(1/2)(A + A2-B2 ) ] . (4.5.5) we find, ∫0 2π dθ1 ln(s212) = ∫0 2π dθ1 ln[a12 + a22 - 2a1a2cos(θ1)] A = a 12 + a22 B = 2a 1a2 A2-B2 = (a12 +a22)2 - 4a12a22 = (a12 -a22)2 => A2-B2 = (a22 -a12) > 0 since a 2 > a1 so Chapter 4: Transmission Line Equations 142 ∫0 2π dθ1 ln(s212) = 2π ln[(1/2)( a 12 + a22 + (a22 -a12) ) = 2πln(a22) . Similarly ∫0 2π dθ2 ln(s122) = 2π ln[(1/2)( a 12 + a22 + (a22 -a12) ) = 2πln(a22) = same as above . The other two integrals are, ∫0 2π dθ1 ln(s112) = ∫0 2π dθ1 ln[2a12(1 - cos(θ 1))] A = B = 2a 12 = 2π ln[(1/2)2a 12] = 2πln(a12) ∫0 2π dθ2 ln(s222) = ∫0 2π dθ1 ln[2a22(1 - cos(θ 2))] A = B = 2a 22 = 2π ln[(1/2)2a 22] = 2πln(a22) . To summarize: ∫0 2π dθ1 ln(s212) = 2πln(a22) ∫0 2π dθ1 ln(s112) = 2π ln(a12) ∫0 2π dθ2 ln(s222) = 2π ln(a22) ∫0 2π dθ2 ln(s122) = 2π ln(a22) . (4.6.2) Then from (4.5.3) we find K = 1/(2 π) { ∫0 2π dθ1 [ln(s212) - ln(s112)] - ∫0 2π dθ2 [ln(s222) - ln(s122)] } = ln(a 22) - ln(a 12) - ln(a 22) + ln(a22) = ln(a 22/a12) = 2 ln(a 2/a1) . (4.6.3) The coaxial transmission line parameters are then given by, C = 4 πε d/K = 2πε d / ln(a2/a1) // centered coaxial G = 4 πσd/K = 2 πσd / ln(a2/a1) Le = (μd/4π)K = (μ d/2π) ln(a2/a1) Z0 = (K /εrel ) 30Ω = (1/ εrel ) ln(a2/a1) 60Ω (4.6.4) We verify the C and L e parameters from http://en.wikipedia.org/wiki/Coaxial_cable , Chapter 4: Transmission Line Equations 143 Fig 4.8 To verify the Z 0 value, first recall that (the positiv e square root is implied here) ch-1x = ln[x + x2-1 ] // Spiegel identity 8.56, valid for x ≥ +1 (4.6.5) If we set x ≡ 1 2 ( a b + b a ) = 1 2 a2+b2 ab and we assume a > 0 and b > 0, then x2 - 1 = 1 4 (a2+b2)2 a2b2 - 1 = 14 { (a2+b2)2 - 4a2b2 a2b2 } = 14 (a2-b2)2 a2b2 => x2-1 = 1 2 |b2-a2| ab = sign(b-a) 1 2 (b a - a b ) => x + x2-1 = 1 2 (a b + b a ) + sign(b-a) 12 (b a - a b ) = ⎩⎨⎧ b/a b ≥ a a/b a ≥ b => ln [x + x2-1 ] = ln [ ⎩⎨⎧ b/a b ≥ a a/b a ≥ b ] = sign(b-a) ln(b/a) . Thus we have shown that (note that both sides are invariant under a ↔ b ) ch-1[1 2 (a b + b a )] = sign(b-a)ln b a . a > 0 and b > 0 (4.6.6) With this rather elaborate fact, and si nce we have b > a, we can rewrite Z 0 above as Z0 = (1/εrel ) ch-1[1 2 (a b + b a )] 60 Ω . (4.6.7) Again we quote from reference RDE page 29-24 Chapter 4: Transmission Line Equations 144 Fig 4.9 In our centered case c = 0 so U = (1/2)(D/d+d/D) and we have agreement. The full off-center result is derived later in Chapter 6, equation (6.3.15). Comment: In the examples of Sections 4.5 and 4.6, the current distributions in the involved round wires are axially symmetric. Therefore all the results of Chap ter 2 apply. In particular, Chapter 2 calculates the surface impedance Z s for a round wire in complete detail, including its limits for large and small ω. For example, at low frequency for a wire of radius a, Z s(ω) = 1 σπa2 + jω μ 8π = Rs + jωLs // low frequency limit (2.4.12) and one sees that R s is the expected DC resistance (C.1.1) and L s is the internal impedance L i as computed in (C.3.5). Having presented our two Examples, we now resume development of the transmission line equations. 4.7 Computation of A z due to one conductor of a transmission line In summary box (1.5.23) we state the following Helmholtz integral for the vector potential arising from currents in a set of conductors, A (x,ω) = 1 4π Σi∫μiJi(x',ω) e-jβR R d V ' (1.5.23) where the sum Σi is over the conductors and μi is the permeability of conductor i. In our transmission line context, and as discussed in Chapter 3, the dominant current is in the z (longitudinal) direction, while transverse currents are very small. For example, in the estimate of Section 3.7 (s) we found that J r/Jz < 1.4 x 10-4 below 10 GHz and J r/Jz < 4.6 x 10-8 at low frequency. Looking at the above Helmholtz solution to th e Helmholtz equation, if we neglect these transverse currents, we are then in effect neglecting the transverse components of A, and this is what we shall do from now on: Chapter 4: Transmission Line Equations 145 Fact : The transverse components A x and Ay can be neglected so that A = Azz^. (Appendix M) (4.7.1) Our starting point then is the following expression for the potential A z at some arbitrary point x in the dielectric due to conductor C1 of a transmission line, Az1(x,ω) = μ1 4π ∫ C1 Jz1(x',y',z',ω) e-jβdR R dx'dy'dz' . R = | x - x'| (4.7.2) Here the point x' = (x',y',z') runs over the volume of C1 and R is the distance between the observation point x in the dielectric and the integration point x'. Parameter β is for the dielectric while μ1 is for the conductor. In the analogous φ solution (4.1.1) everything has the same form as (4.7.2) but in (4.1.1) the charge density exists only on the conductor surface. Nevert heless, we represented that charge density as a volume density, and only later in examples set that volume density to a surface distribution. Thus, the parallel between the φ and the A z analysis is very close, not surprising in light of (1.3.11). Another difference is that for φ the leading factor is 1/(4πξ d) where ξd was the complex dielectric constant of the dielectric. In (4.7.2) this factor is replaced by ( μ1/4π) where μ 1 is the magnetic permeability of the conductor C 1. We next make the same assumption of separation of variables to write J z1(x,y,z) = b 1(x,y) i1(z) A/m2 1/m2 A (4.7.3) where i 1 is scaled such that ∫C1 dx dy b1( x , y ) = 1 . ( 4 . 7 . 4 ) Here b1(x,y) describes the distribution of the current density across the conductor C 1 cross section. At DC this density is a uniform constant, but at higher ω the density becomes non-uniform in two ways. First, it becomes concentrated away from the central region due to the skin effect. Second it is non-uniform in that it tends to concentrate on the portion of conductor C 1 which is closest to conductor C 2. In the corresponding equation ρ1(x,y,z) = α1(x,y) q1(z) of (4.1.2), α1(x,y) exists only on the conductor surface, and is generally non-uniform in the second sense noted above for b 1(x,y). As before, we can now interpret i 1(z) as the total current in C1 at z. Again assuming that there is no net superposed radiating antenna curre nt, we have equal and opposite curre nts in the two conductors so the line is a balanced line, and then i2(z) = - i1(z) = -i(z) . (4.7.5) Chapter 4: Transmission Line Equations 146 This then leads to Az1(x,y,z) = μ1 4π ∫-∞ ∞ dz' i(z') ∫C1 dx' dy' b 1(x',y') e-jβdR R . (4.7.6) Note : In the ω domain, J z1(x,y,z) is complex with a position-dependent phase, as for example in the plot of Ez = Jz/σ shown in Fig 2.8. Thus, b 1(x,y) is complex and has a position-dependent phase. This does not stop us from allocating the phase between the two terms in (4.7.3) so that the integral of b 1(x,y) is unity as in (4.7.4). On e ends up then with i 1(z) having some phase that in general is non-zero. Comments regarding μ This is a subtle subject and is not discus sed in King's trans mission line theory book. If the conductor C 1 and dielectric have the same permeability so that μ1 = μd, then there exists no "magnetic boundary" between the conducto r and dielectric. The solution (4.7.2) is then smooth at this boundary, and so A z1(x,y,z) "naturally" satisfies these two boundary conditions, Az1(x+) = Az1(x- ) (1/μd)∂nAz1(x+) = (1/μ 1) ∂nAz1(x- ) ( 4 . 7 . 7 ) where x+ is just outside the conductor surface and x- is just inside. The second equation here is just (1.1.46) in the case that there is no free surface current K free flowing on the boundary, and indeed in our example at hand there is no such free surface current. Since we have assumed that μd = μ1, this second boundary condition just says ∂nAz1(x+) = ∂nAz1(x+). Since there is no magnetic boundary at the conductor/dielectric interface, the so lution (4.7.2) is continuous and all its derivatives are also continuous at the boundary, since nothing special happens at that boundary. Thus, the Helmholtz integral solution provides the whole solution for A z1 since it meets both "boundary conditions" at this pseudo boundary. If on the other hand we have μ1 ≠ μd, then there is a magnetic boundary between conductor and dielectric which we have to worry about. In this case, (4.7.2) cannot possibly satisfy the second boundary condition of (4.7.7) since, as already noted, the A z1 of (4.7.2) satisfies ∂ nAz1(x+) = ∂nAz1(x+). Thus, in this case (4.7.2) is not the full solution for A z1. One must add a homogeneous Helmholtz equation solution to (4.7.2) in order to have a proper solution for A z1 that satisfies both equations in (4.7.7). It turns out that the correct total A z1 solution can be generated by adding a certain fictitious surface current term to μ1Jz1 in (4.7.2). Since such a surface curre nt vanishes on both sides of the boundary between μ and μ1, the Helmholtz solution due just to this su rface current term is in fact a homogeneous solution to the Helmholtz equation in both the condu ctor and dielectric regions, away from that boundary. It turns out moreover that the correct fictitious surface current to add is in fact the magnetization surface current J m which is created at the boundary between μ d ≠ μ1. Adding this surface current is just a "trick" in order to generate the correct homogeneous adder solution so that the resulting total A z1 satisfies both boundary conditions in (4.7.7). Formally speaking, the J i appearing in (1.5.4) and then J z1 in (4.7.2) should not include such magnetization currents since this J is really the J in Maxwell's equation curl H = Chapter 4: Transmission Line Equations 147 ∂tD + J, and this J does not include magnetization currents -- it includes only normal conduction currents. In our current Chapter 4, we want (4.7.2) to represent the complete solution for A z1 and for that reason we must restrict our analysis to the situa tion where dielectric and all conductors have the same permeability which we shall just call μ. In practice, one normally has μd = μ1 = μ0. In order to handle the more general case of μ d ≠ μ1, we have to deal with the inhomoge neous adder solutions or equivalently with the abovementioned fictitious surface current, a nd this complicates our analysis which is already quite complicated. So, for the moment, we now make the same assumption made by King and other authors: Fact : From now on, conductors and dielectric must have the same permeability μ d. (4.7.8) After fully developing this special case, we shall then extend the theory in Section 4.12 to allow for μ d ≠ μ1. Appendix G shows for the round wire how the inhomogeneous adder solution is found and how it then causes the boundary conditions (4.7.7) to be met when μ1 ≠ μd. Appendix B shows how the addition of a fictitious surface current term μ0Jm provides an alternate and simpler solution to the same problem of meeting boundary conditions (4.7.7) when μ1 ≠ μd. It then shows exactly how this works in the special case of a round wire. Having now mentioned that the Helmholtz integral mi ght not provide a total solution, the reader might fairly ask why it is that the Helmholtz integral solution φ 1(x,ω) of (4.1.1) provides a complete and viable solution to the φ Helmholtz equation, given th at in general the conductor (ε 1) and dielectric ( ε) have different ε values, so there should be an "electric boundary" where ε meets ε1. The reason is that, according to (1.1.47), the boundary condition corr esponding to the second line of (4.7.7) reads [ε1En(x+) - εEn(x+)] = n free(x) . Since we are neglecting transverse A components as stated in (4.7.1), and since our notation ∂ n indicates a normal conductor derivative which is transverse (to z), we have E = - grad φ - ∂ tA => E n = -∂nφ ( 4 . 7 . 9 ) so we have then this set of boundary conditions for φ 1, φ 1(x+) = φ1(x-) [ε1∂nφ1(x+) - ε∂nφ1(x-)] = nfree(x) . ( 4 . 7 . 1 0 ) These look a bit like (4.7.7) for A z1. The big difference is that in this case there does exist a free surface charge n free and it simply adjusts itself to make (4.7.10) be true. Thus, the Helmholtz integral (4.1.1) does in fact meet the required electrical boundary conditions without the need for a homogeneous solution adder term. A less formal way to state this is that, in the electrical case, we can regard the surface charge Chapter 4: Transmission Line Equations 148 as in fact lying on the dielectric si de of the boundary, and then the boundary is of no interest in our problem of analyzing fields in the dielectric. 4.8 Computation of potential A z due to both conductors of a transmission line Let us now write the potential at an arbitrary point x in the dielectric due to both conductors C 1 and C2. We accept the requirement of (4.7.8) and re quire that all conductors have the same μ = μd as the dielectric, so then μ1 = μ and μ2 = μ. Then, Az12(x) = Az1(x) + Az2(x) = μd 4π ∫-∞ ∞ dz' i(z') { ∫C1 dx1' dy1' b1(x1',y1') e-jβdR1 R1 – ∫C2 dx2' dy2' b2(x2',y2') e-jβdR2 R2 } R 12 = (x-x1')2 + (y-y1')2 + (z-z')2 = s12 + (z-z')2 s 12 = (x-x1')2 + (y-y1')2 (4.8.1) R 22 = (x-x2')2 + (y-y2')2 + (z-z')2 = s22 + (z-z')2 s22 = (x-x2')2 + (y-y2') . The picture going with the above equation is iden tical to Fig 4.1 below (4.2.1) except the points x1' and x2' can be in the interior of the conductors, not just on the boundary of the conductors. 4.9 Transmission Line Limit Revisited In Section 4.3 we discussed the so-called trans mission line limit of small β in the context of the scalar potential φ. We could (but won't) repeat the discussion verbatim here making th e following substitutions: q(z) → i(z) αi → bi φ12 → Az12 1 4πξ → μ 4π . The conclusion is that in the transmission line limit (small β, long wavelength λ = 2π/β) we may write Az12(x) = μd 4π i(z) ∫-∞ ∞ dz'{ ∫C1 dx1' dy1' b1(x1',y1') 1 R1 – ∫C2 dx2' dy2' b2(x2',y2') 1 R2 } (4.9.1) which is analogous to (4.3.10). Also, in analogy with (4.3.8) we write i(z) = i(0) e -jβdz // i(z,t) = i(0,0) ej(ωt-βdz) (4.9.2) 4.10 General Calculation of W(z) As before, we now intr oduce the two new points x1 and x2. The point x1 lies on C 1 in the z = z plane, while x2 lies on C 2 in this same plane. We then evaluate A z12(x) at x = x1 and subtract from that A z12(x) at x = x2 and in this way we obtain the A z potential difference between the surfaces of the two conductors at z = z which we shall call W(z). Recall, Chapter 4: Transmission Line Equations 149 Fact 5: On each conductor boundary, A z ≈ constant in the extreme or strong skin effect regimes. (3.7.20) Thus, assuming the small δ regime and treating A z ≈ constant as an equality, the A z potential difference will be independent of the locations of x2 and x1 as long as they are on their respective surfaces and both have z = z. For this reason, the A z potential difference is a function only of z. Thus we write, using two copies of (4.9.1), W(z) ≡ Az12(x1) - Az12(x2) = μd 4π i(z) ∫-∞ ∞ dz' { ∫C1 dx1' dy1' b1(x1',y1') 1 R11 – ∫C2 dx2' dy2' b2(x2',y2') 1 R12 } – μd 4π i(z) ∫-∞ ∞ dz' { ∫C1 dx1' dy1' b1(x1',y1') 1 R21 – ∫C2 dx2' dy2' b2(x2',y2') 1 R22 } (4.10.1) where R112 = (x1-x1')2 + (y1-y1')2 + (z-z')2 = s112 + (z-z')2 s 112 = (x1-x1')2 + (y1-y1')2 R122 = (x1-x2')2 + (y1-y2')2 + (z-z')2 = s122 + (z-z')2 s 122 = (x1-x2')2 + (y1-y2')2 R222 = (x2-x2')2 + (y2-y2')2 + (z-z')2 = s222 + (z-z')2 s 222 = (x2-x2')2 + (y2-y2')2 R212 = (x2-x1')2 + (y2-y1')2 + (z-z')2 = s212 + (z-z')2 s 212 = (x2-x1')2 + (y2-y1')2 . (4.10.2) The picture going with the above equation is identical to Fig 4.2 below (4.4.2) except, once again, the points x1' and x2' can be in the interior of the conductors, not just on the surface of the conductors. Also, we replace the figure's double arrow label V(z) with W(z). We then reorder the four terms to get W ( z ) ( 4 . 1 0 . 3 ) = μ d 4π i(z) ∫-∞ ∞ dz' { ∫C1 dx1' dy1' b1(x1',y1')( 1 R11 - 1 R21 ) - ∫C2 dx2' dy2' b2(x2',y2') (1 R12 - 1 R22 ) } . The dz' integrals are the same as those done in Section 4.4 and we then arrive at W(z) = i(z) μd 4π { ∫C1 dx1' dy1' b1(x1',y1') ln(s212/s112) - ∫C2 dx2' dy2' b2(x2',y2') ln(s222/s122) } s212 = (x2-x1')2 + (y2-y1')2 s 222 = (x2-x2')2 + (y2-y2')2 (4.10.4) s112 = (x1-x1')2 + (y1-y1')2 s 122 = (x1-x2')2 + (y1-y2')2 which is analogous to (4.4.6) for V(z). The corresponding drawing is analogous to Fig 4.3 where, once again, the integration points x1' and x2' are inside the conductor : Chapter 4: Transmission Line Equations 150 Fig 4.10 Equation (4.10.4) expresses the A z potential between the two transmission line conductors at some plane z in terms of the current distributions b i within the conductors. Now, the Stokes theorem applied to B = curl A says curl A = B ⇔ ∫{ A • ds = ∫S B • dS . (1.1.39) Consider the red loop shown in this top view of the two transmission line conductors. The loop is intended to have a tiny width dz, and the top view ob scures the fact that each conductor has an arbitrary cross section. The loop makes contact with the points x1 and x2 shown in the previous figure, Fig 4.11 Since we neglect any transverse components of A, the Stokes theorem says [Az1(top) - A z2(bottom) ] dz = [ magnetic flux through red loop] = ∫S B • dS . (4.10.5) If we regard the two short dz length conductor pieces as forming a tiny "inductor", closed on the ends by the vertical red lines, we can use this definition of in ductance to compute the inductance of that inductor: Chapter 4: Transmission Line Equations 151 [magnetic flux through red loop] = (L edz) i(z) . (4.10.6) Here (Ledz) is the inductance of our tiny loop, so L e is the transmission line inductance per unit length. We know (as in Appendix C) that there will be magnetic flux inside the conductors as well as between them, and for that reason L e as defined here only accounts for the "external" inductance of the transmission line, again see Appendix C. Since [A z1(top) - A z2(bottom) ] = W(z) according to (4.10.1) , we may combine (4.10.5) and (4.10.6) to obtain W(z) = L e i ( z ) . ( 4 . 1 0 . 7 ) Therefore from (4.10.4) we have found that L e = W(z) i(z) = μd 4π KL ( 4 . 1 0 . 8 ) where K L is the following dimensionless real number, KL ≡ ∫C1 dx1' dy1' b1(x1',y1') ln(s212/s112) - ∫C2 dx2' dy2' b2(x2',y2') ln(s222/s122) . (4.10.9) This number is reminiscent of the number K obtained in Section 4.4, K ≡ ∫C1 dx1' dy1' α1(x1',y1') ln(s212/s112) - ∫C2 dx2' dy2' α2(x2',y2') ln(s222/s122) . ( 4.4.8) In the next section it will be shown that these two dimensionless numbers are exactly the same. 4.11 The Classical Transmission Line Equations The results of the previous sections of this chapter may be succinctly summarized as: ( 4 . 1 1 . 1 ) 1 C' = V(z) q(z) = 1 4πξd K (4.4.7) Le = W(z) i(z) = μd 4π KL (4.10.8) K ≡ ∫C1 dx1' dy1' α1(x1',y1') ln(s212/s112) - ∫C2 dx2' dy2' α2(x2',y2') ln(s222/s122) (4.4.8) KL ≡ ∫C1 dx1' dy1' b1(x1',y1') ln(s212/s112) - ∫C2 dx2' dy2' b2(x2',y2') ln(s222/s122) (4.10.9) Chapter 4: Transmission Line Equations 152 Notice that we have made no assumptions whatsoever about the cross-sectional shape of the transmission line. We have only assumed that the transverse dimensions are small compared to the wavelength λ that corresponds to β -- this was the transmission line limit. (a) Initial Processing There are s everal equations from Chapter 1 we shall now press into service: E = - grad φ - ∂tA (1.3.1) div A = - μdεd ∂tφ - μσφ . // the King gauge (1.3.18) In the frequency domain these become, E = - grad φ - jωA div A = - j (β d2/ω) φ . // the King gauge, see (1.5.1a) and (1.5.5) re βd2 (4.11.2) According to the Fact stat ed in (4.7.1), potential A has only component A z, so these equations become Ez(x) = - ∂zφ(x) - jω Az(x) ∂zAz(x) = - j (β d2/ω) φ(x) . (4.11.3) However, as was shown at the end of Step 1 below (3.7.8), the second line of (4.11.3) can only be justified in the strong or extreme skin effect regimes, and we continue then to assume our transmission line is operating at sufficiently high ω to be in the small δ regime. The potentials in the above equations are those due to both conductors and were denoted as φ 12 and Az12 in the previous sections. We then rewrite the above as E z(x) = - ∂zφ12(x) - jω Az12(x) ( 4 . 1 1 . 4 a ) ∂zAz12(x) = - j (β d2/ω)φ12(x) . (4.11.4b) Recall now the conductor-surface-located points x1 and x2 as shown for example in Fig 4.10. If we evaluate each of the above equations at x = x1 and then x = x2 and then subtract, we get Ez(x1) - Ez(x2) = -∂z[φ12(x1) - φ12(x2)] - jω[Az12(x1) - Az12(x2)] (4.11.5a) ∂z[Az12(x1) - Az12(x2)] = - j (β2/ω)[φ12(x1) - φ12(x2)] . (4.11.5b) Then using these definitions (again, we are assuming the strong or extreme skin depth regime) V(z) ≡ φ 12(x1) - φ12(x2) (4.4.1) W(z) ≡ Az12(x1) - Az12(x2) (4.10.1) we may rewrite (4.11.5) in this simple manner, Chapter 4: Transmission Line Equations 153 Ez(x1) - Ez(x2) = - ∂zV - jωW ( 4 . 1 1 . 6 a ) ∂zW = - j (βd2/ω) V . ( 4 . 1 1 . 6 b ) The quantity E z(x1) is the longitudinal electric field at point x1 on the surface of conductor C 1. It is related to the conductor's at-the -surface current density by J z(x1) = σ Ez(x1). If the conductor were "perfect", we would have σ = ∞ and Ez(x1) = 0, but real conductors are not perfect. However, since we are assuming the strong or extreme skin effect all along here in our analysis, we do know that E z(x1) and Ez(x2) are very small. (b) Averaging Repair and the Transmission Line Equations Our theor y now has an inconsistency which needs to be fixed. We know that for a general transmission line operating at ω > 0, the current density J z inside the conductors will not be uniformly distributed. It will be larger in the conductor region closest to the other conductor. This "proximity effect" is discussed in Appendix P from an eddy current point of view, see Fig P.13 for an example. The J z current non-uniformity can be very dram atic as for example in a transmission line having this cross section, where J z will be large near the gap and small far from the gap: Fig 4.12 Since Jz is non-uniform in each conductor, so is E z, and so we expect E z(x1) to be a strong function of the point x1 on the perimeter of C 1, certainly for the above cross section example. This means that the left side of (4.11.6a) is a function of x1 = (x1,y1,z) and x2 = (x2,y2,z) whereas the right side in our theory is a function only of z. To remedy this inconsistency, we now have to think of V and W as having very slight dependence on x 1 and x2 which we generally ignore, but which we must face up to in (4.11.6a). In reality we have V( x1,x2) and W( x1,x2). This is a manifestation of the fact that in reality φ ≈ constant and A z ≈ constant on the boundaries (with ≈ and not = ). In the extreme skin effect regime (think a very good conductor), the left side of (4.11.6a) can be a violent function of x1 and x2 as in the case of the above figure, but the left side is always very small, even where it is largest, and its variation can be accommodated by the right side of (4.11.6a) which is the difference of large-valued functions which vary only slightly with x1 and x2. So first rewrite (4.11.6a) as Ez(x1) - Ez(x2) = - ∂z V(x1,x2) - jω W(x1,x2) . (4.11.6a)' Backing up another step, we write out of (4.11.4a) for the two perimeter points x 1 and x2, (1/σ)J z(x1) = Ez(x1) = - ∂zφ12(x1) - jω Az12(x1) x1 on perimeter of C 1 (1/σ)Jz(x2) = Ez(x2) = - ∂zφ12(x2) - jω Az12(x2) x2 on perimeter of C 2 (4.11.4a)' Chapter 4: Transmission Line Equations 154 Calling the perimeter distances of the conductors P 1 and P2, we then average each of these equations around its appropriate perimeter. Apply (1/P 1) ∫C1 ds1 to the first equation and (1/P 2) ∫C1 ds2 to the second to get [ ds 1 is a distance element along the perimeter of C 1 ] , (1/σ)<Jz(x1)>C1 = <Ez(x1) >C1 = - ∂z<φ12(x1) >C1 - jω<Az12(x1) >C1 (1/σ)<Jz(x2)>C2 = <Ez(x2) >C2 = - ∂z<φ12(x2) >C2 - jω<Az12(x2) >C2 . Subtract the second line from the first to get, [<E z(x1) >C1 - <Ez(x2) >C2] = - ∂z[<φ12(x1) >C1 - <φ12(x2) >C2] - jω [<Az12(x1) >C1 - <Az12(x2) >C2 ] . We now redefine V and W to be the averages appearing in these equations, along with E z1 and Ez2 : Ez1(z) ≡ <Ez(x1) >C1 = (1/P 1) ∫C1 ds1 Ez(x1) Ez2(z) ≡ <Ez(x2) >C2 = (1/P2) ∫C2 ds2 Ez(x2) V(z) ≡ <φ12(x1) >C1 - <φ12(x2) >C2 = <V( x1,x2)>C1,C2 W(z) ≡ <Az12(x1) >C1 - <Az12(x2) >C2 = <W(x 1,x2)>C1,C2 (4.11.7) with this result [Ez1(z) - Ez2(z)] = - ∂z V(z) - jω W ( z ) . ( 4 . 1 1 . 8 ) Meanwhile, the surface impedances on C 1 and C2 are defined by (see C.2.1) , E z1(x1) = Zs1(x1) i1(z) Ez2(x2) = Zs2(x2) i2( z ) (C.2.1) which we average in the same way to obtain Ez1(z) = Z s1 i1(z) Z s1 ≡ (1/P1) ∫C1 ds1 Zs1(x1) Ez2(z) = Z s2 i2(z) Z s2 ≡ (1/P2) ∫C2 ds2 Zs2(x2) . (4.11.9) There will be some location on C 1 where E z1(x1) and thus Z s1(x1) will be maximal (for example on the walls of the gap in Fig 4.12). Referring to this value as Z s1,max we can define p 1 ≡ (Zs1/Zs1,max )P1 p2 ≡ (Zs2/Zs2,max )P2 ( 4 . 1 1 . 1 0 ) Chapter 4: Transmission Line Equations 155 where p1 is the effective length of the "active perimeter" of C 1. This then provides a crude model for the symbol p which appears in (2.5.1) and Fig 2.16 which we replicate here, Fat twinlead Fig 2.16 Then using i(z) = i 1(z) = -i2(z) and (4.11.9), rewrite (4.11.8) and (4.11.6b) as [Z s1 + Zs2] i(z) = - ∂z V(z) - jω W(z) ∂zW(z) = - j ( βd2/ω) V ( z ) ( 4 . 1 1 . 1 1 ) where the second equation above is the < > C1,C2 average of (4.11.6b). Continuing this repair effort, we back up to box (4.11.1) and write V( x1,x2) = q(z) / C'( x1,x2) = q(z) [ 1 4πξd K(x1,x2) ] W( x1,x2) = i(z) L e(x1,x2) = i(z) [ μ 4π KL(x1,x2)] (4.11.12) which we average in the same way to get V(z) = q(z) 1 C' W(z) = i(z) L e 1 C' ≡ (1/P1) ∫C1 ds1 (1/P2)∫C2 ds2 1 C'(x1,x2) = < 1 C'(x1,x2) >C1,C2 Le ≡ (1/P1) ∫C1 ds1 (1/P2)∫C2 ds2 Le(x1,x2) = < L e(x1,x2)>C1,C2 . (4.11.13) The "constants" K and K L in (4.11.1) are similarly replaced with their <> C1,C2 averages. Inserting the equations on the first line of (4.11.13) into (4.11.11) we get (Z s1 + Zs2) i(z) = - ∂zV(z) - jω Le i(z) Le ∂z i(z) = - j ( βd2/ω) V(z) which we then rearrange as ∂ zV(z) = - [ Z s1+ Zs1+ jωLe] i(z) ∂z i(z) = - [ j βd2/(ωLe) ] V ( z ) . ( 4 . 1 1 . 1 4 a ) Chapter 4: Transmission Line Equations 156 These are the classical transmission line equations . They are usually written in this form: [ ∂/∂z = d/dz] dV(z) dz = - z i(z) di(z) dz = - y V(z) with z = R + jωL y = G +jω C . (4.11.14b) Note : We have been using bold notation only for vectors, and we now break that guideline by bolding these complex quantities z and y. Our purpose for this bolding is to distinguish them from Cartesian coordinates z and y which typically appear in the same problem. In King's books, all complex parameters are put in bold font, but we do this only for z and y. If one applies ∂ z to either of the above equations and then uses the other, one obtains the corresponding wave equations (but in the ω domain, so Helmholtz equations), d2V(z) dz2 - zy V(z) = 0 d2i(z) dz2 - zy i(z) = 0 . (4.11.15) In the discussion below, we shall no longer menti on the averaging process, but it should be understood that for closely spaced conductors the symbols Z s1, Zs1, Le, C', K, K L, V, W are the perimeter-averaged values discussed above. For widely spaced conductors, J z is roughly uniform over the conductor cross sections and perimeters and the averaging process is not needed. Jumping the gun a bit, if we assume now a traveling-wave z dependence e j(ωt-kz) for both V(z) and i(z), where k is the wave's (possibly complex) wavenumber, then ∂z → -jk and the transmission line equations become -jk V(z) = - z i(z) -jk i(z) = - y V(z) or -jk = - z i(z)/V(z) -jk = - y V(z)/i(z) . Equating these last two expressions gives - z i(z)/V(z) = - y V(z)/i(z) => z/y = [V(z)/i(z)] 2 and we then have, Z 0 ≡ V(z)/i(z) = z y = R + jωL G + jωC (4.11.16) where by definition Z 0 is the characteristic impedance of the transmission line. The quantities z and y are called the transmission line impedance and admittance , and the four numbers R,L,G,C are defined to be the appropriate real and imaginary parts. Comparing (4.11.14a) and (4.11.14b), we may therefore conclude that: Chapter 4: Transmission Line Equations 157 z = R + jωL = Zs1 + Zs2 + jωLe ( 4 . 1 1 . 1 7 ) y = G + jω C = jβ d2/(ωLe) ( 4 . 1 1 . 1 8 ) where Z s1, Zs2 and Le are the averages shown in (4.11. 9) and (4.11.13). The expression for z seems quite reasonable since ωLe = XL = inductive reactance, but the expression for y seems a bit unusual. This is because we still have more work to do. There is one more equation we have not yet utilized . Recall from Chapter 1 the integral form of the equation of continuity, which in the frequency domain takes this form, div J = - jωρ ⇔ -jω [ ∫V ρ dV] = ∫S J • dS . (1.1.35) We now apply this to a Gaussian box (blue) whos e faces have the same shap e as the conductor cross section but are slightly larger than that cross sec tion so as to include the conductor surface charge : Fig 4.13 Ignoring transverse dielectric current out the radial sides of the box (since dz is tiny), we get -jω [q(z)dz] = i(z+dz) - i(z) = total current flowing out of the box which then says ∂ z i(z) = -jωq(z) . (4.11.19) Jumping the gun again, if we again use ∂ z → -jk with k = ( ω/v), we arrive at the intuitive relation i(z) = q(z) v ( 4 . 1 1 . 1 9 a ) which just says the charge per unit length is [in e ffect, see D.9 (c)] traveling down the line at phase velocity v. In the lossless case v = v d (dielectric speed of light), whereas more generally v is complex. From summary box (4.11.1) recall that q(z) = C' V(z) so we get from (4.11.19), ∂ z i(z) = - [ j ωC ' ] V ( z ) . ( 4 . 1 1 . 2 0 ) Comparing with the second equation of (4.11.14), Chapter 4: Transmission Line Equations 158 ∂z i(z) = - [ j βd2/(ωLe)] V ( z ) , (4.11.14) we get the following identity, - [ jβ d2/(ωLe)] = - [ jωC'] or L eC' = βd2/ω2 = μdξd . // see (1.5.1a) regarding βd2 (4.11.21) Then we can write (4.11.18) as y = G + jω C = jβ d2/(ωLe) = j (βd2/ω2) (ω/Le) = j (LeC') (ω/Le) = jωC' . (4.11.22) Thus, the line capacitance C is the real part of complex capacitance, C = Re(C'), and G = - ω Im(C'). (c) Digression on the meaning of C' Back in Section 1.5 (c) we discussed the fact that n c = (ξd/εd) ns which relates actual surface charge n s to the adjusted surface charge density n c which allows for dielectric leakage. This relationship (1.5.17) was derived in two different ways. As noted in Comment 3 at the start of Section 4.1, and looking at (4.1.1) and (4.1.2), one sees that the linear charge density q(z) which appears in all our equa tions is in fact related to nc and not n s, so we temporarily shall refer to q(z) as q c(z). Then q c(z) = ∫nc dxdy = an integral over the conductor surface for length dz. The actual charge on the surface of this piece of conductor is q s(z) ≡ ∫ns dxdy and therefore q c/qs = nc/ns = (ξd/εd). The capacitance C per unit length of our transmission line is defined by q s = C V(z) . The complex capacitance C', which includes the effect of dielectric leakage current, is defined by q c = C' V(z). Therefore C'/C = q c/qs = (ξd/εd) . ( 4 . 1 1 . 2 3 ) and so then from (4.11.22), (4.11.23) and (1.5.1a), y = G + jω C = jωC' = jω (ξ d/εd)C = jω (1 - jσd/εdω)C = jωC + (σd/εd)C (4.11.24) so that G = ( σ d/εd) C . ( 4 . 1 1 . 2 5 ) We saw an example of (4.11.23) in (1.5.19) for a para llel plate capacitor, and more generally in (4.4.10). We may now rewrite the first equation in summary box (4.11.1) as 1 C' = 1 4πξd K => 1 C = 1 4πεd K . (4.11.26) Chapter 4: Transmission Line Equations 159 Next, combining (4.11.21) and (4.11.23) we find that LeC' = μdξd ( 4 . 1 1 . 2 7 ) L eC = μdεd = 1/vd2 ( 4 . 1 1 . 2 8 ) where v d is the speed of light in the dielectric. Now the second equation in (4.11.1) says that Le = μd 4π KL . ( 4 . 1 1 . 2 9 ) Inserting (4.11.29) for L e and (4.11.26) for C' into (4.11.27) gives (μd 4π KL ) (4πξd/K) = μdξd or KL = K . ( 4 . 1 1 . 3 0 ) This is a remarkable connection between our two seemingly unrelated constants K and K L, K ≡ ∫C1 dx1' dy1' α1(x1',y1') ln(s212/s112) - ∫C2 dx2' dy2' α2(x2',y2') ln(s222/s122) (4.4.8) KL ≡ ∫C1 dx1' dy1' b1(x1',y1') ln(s212/s112) - ∫C2 dx2' dy2' b2(x2',y2') ln(s222/s122) . (4.10.9) Since K involves a peripheral line integral of surface charge densities αi whereas K L involves a full cross sectional area integral of the current densities b i, it seems unlikely these integrals would be equal, but they are equal. (d) An example of K = K L The equality even seems unlikely in a case with sy mmetric densities on round wires, so let's do a check using our Section 4.5 example with widely-spaced r ound wires of unequal diameters. The first thing we need is a new picture to display the "kinematics" of the K L integral ( since densities are symmetric, one should regard this picture as having b much larger than shown relative to a 1 and a2), Chapter 4: Transmission Line Equations 160 Fig 4.14 As before, we read off the four distances of interest using the law of cosines. The new distances are all different than they were before since x 1' and x2' are now each integrated ov er their respective disks instead of the bounding circles. s 212 = r12 + (b-a1)2 - 2 r1(b-a1) cos(θ1) s112 = r12 + a12 - 2 r1 a1 cos(θ1) s222 = r22 + a22 + 2 r2 a2 cos(θ2) s122 = r22 + (b-a2)2 + 2 r2(b-a2) cos(θ2) . The integration rule is still ∫0 2π dθ ln (A ± Bcos θ) = 2π ln[(1/2)(A + A2-B2 )] . (4.5.5) The first integral is: ∫0 2π dθ1 ln(s212) = ∫0 2π dθ1ln([r12 + (b-a1)2 - 2 r1(b-a1) cos(θ1)] A = r 12 + (b-a1)2 B = 2 r 1(b-a1) A2-B2 = [r12 + (b-a1)2]2 - 4 r12(b-a1)2 = [r12 - (b-a1)2]2 => A2-B2 = (b-a1)2- r12 > 0 b >> a 1 => ∫0 2π dθ1 ln(s212) = 2π ln[(1/2)( r12 + (b-a1)2 + (b-a1)2 - r12 ) = 2π ln[(b-a1)2] But this integral is the same as before! The s 112 integral is obtained from the above with b-a 1→a1 ∫0 2π dθ1 ln(s112) = 2π ln(a12) which is also the same as before. The other two integrals are found from 1 → 2. Our integral summary is then exactly the same as (4.5.6), Chapter 4: Transmission Line Equations 161 ∫0 2π dθ1 ln(s212) = 2π ln[(b-a1)2] ∫0 2π dθ1 ln(s112) = 2π ln(a12) ∫0 2π dθ2 ln(s222) = 2π ln(a22) ∫0 2π dθ2 ln(s122) = 2π ln[(b-a2)2] . (4.5.6) We now assume that the current densities b i each have radial symmetry ("widely spaced wires") , b1(r1,θ1) = b1(r1) (4.11.31) where b 1(r1) is a completely arbitrary function, with the following normalization of (4.7.4), ∫0 2π dθ1 ∫0 a1 r1dr1 b1(r1) = 1 => ∫0 a1 r1dr1 b1(r1) = 1/2π . (4.11.32) We now proceed to calculate the constant K L KL = ∫0 2π dθ1 ∫0 a1 r1dr1 b1(r1) ln(s212/s112) - ∫0 2π dθ2 r2dr2 b2(r2) ln(s222/s122) = ∫0 a1 r1dr1 b1(r1) ∫0 2π dθ1 ln(s212/s112) - ∫0 a2 r2dr2b2(r2) ∫0 2π dθ2 ln(s222/s122) = 2 π ∫0 a1 r1dr1 b1(r1) [ln[(b-a 1)2]- ln(a12)] - ∫0 a2 r2dr2b2(r2) [ ln[(b-a 2)2] - ln(a22)] = 2 π [ln[(b-a 1)2/a12] ∫0 a1 r1dr1 b1(r1) - 2π [ln[(b-a 2)2/a22] ∫0 a2 r2dr2 b2(r2) = [ln[(b-a 1)2/a12] - [ln[(b-a 2)2/a22] = l n [(b-a1)2(b-a2)2 a12a22 ] = K as obtained in (4.5.7) (4.11.33) and we have then shown K L = K for this particular example. The key fact is that the d θ integrals appear to be functions of r i , but the r i2 terms cancel and so the d θ integrals are independent of r i. Chapter 4: Transmission Line Equations 162 (e) Summary of Results Classical Transmission Line Equations and Parameters (ω domain) (4.11.34) K ≡ ∫C1 dx1' dy1' α1(x1',y1') ln(s212/s112) - ∫C2 dx2' dy2' α2(x2',y2') ln(s222/s122) (4.4.8) KL ≡ ∫C1 dx1' dy1' b1(x1',y1') ln(s212/s112) - ∫C2 dx2' dy2' b2(x2',y2') ln(s222/s122) , (4.10.9) K = K L real and dimensionless (4.11.30) dV(z) dz = - z i(z) ( d dz2 - zy) V(z) = 0 z = R + jωL transmission line equations di(z) dz = - yV(z) (d dz2 - zy) i(z) = 0 y = G +jω C (4.11.14) , (4.11.15) z = Zs1 + Zs2 + jωLe (4.11.17) XL ≡ ωLe , XC ≡ 1/(ωC) y = jω C' = jωC + (σd/εd)C (4.11.24) G = (σ d/εd)C (4.11.25) R = Re(Z s1+ Zs2) L = L e + (1/ω) Im(Zs1+ Zs2) = Le + Li Le = μd 4π K (4.11.29) and (4.11.30) C' = 4πξd/K (4.11.26) C' = (ξd/εd)C (4.11.23) C = 4πεd/K (4.11.26) G = 4πσd/K (4.11.26) + (4.11.25) => G/C = σd/εd LeC' = μdξd (4.11.27) LeC = μdεd = 1/vd2 (4.11.28) Z0 = R + jωL G + jωC = z y (4.11.16) Z0 (large ω) ≈ L C ≈ Le C = (1/4π) K μd/εd = (K/4π) Zm // See comments below λ >> D (4.3.6) assumed transmission line limit where βd = 2π/λ βd2 = μdεdω2 - jωμdσd = ω2μd ( εd - jσd/ω) = ω2μd ξd ξd ≡ εd - jσd/ω . (1.5.1a) F(z) = F(0) e-az e-jbz a ≡ Re( zy ) = Re[ (R+jωL)(G+jωC) ] b = Im( zy ) see (5.3.6) attenuation phase Chapter 4: Transmission Line Equations 163 Comments: 1. In Chapter 2 we computed the surface impedance Z s for a round wire in the case of axially symmetric current and we found that, for large ω, Zs(ω) ≈ 1 σ(2πa)δ ( 1 + j ) (2.4.16) δ ≡ 2/ωμσ = skin depth (2.2.20) so that Zs(ω) ≈ 1 2πa μ 2σ (1+j) ω . (4.11.35) Presumably the result will be Z s(ω) ~ ω for any conductor cro ss section shape. Then L = L e + (1/ω) Im(Zs1+ Zs2) = Le + (stuff) 1/ ω → Le for large ω (4.11.36) For this reason, the high freque ncy characteristic impedance Z 0 can be written as shown in (4.11.34). 2. Conductors have internal inductance L i as well as external inductance L e. In Appendix C.3 (a) we compute the low frequency internal inductance of a round wire to be L i = μ/8π = (μ/μ0) * 50 nH/m . Our Chapter 4 transmission line development makes no mention of L i. This can be traced to Figure 4.11 where only the external magnetic flux is involved. In fact, L i is accounted for in the imaginary part of the surface impedance Z s . For example, we found that for our round wire situation, Zs(ω) = 1 σπa2 + jω μ 8π = Rs + jωLs // low frequency limit (2.4.12) and here one sees that L s = μ 8π = Li . 3. We have assumed that εd and μd are real. If not, the usual adjustment s can be made in (4.11.34) for the interpretations of R,L,G and C. See for example (3.3.4) concerning σ being replaced by σeff if ε has an imaginary part. 4. Apart from the symmetric cases like the examples of Section 4.5 and 4.6, we do not yet have a way to compute K and the transmission line parameters since the charge and current distributions αi and bi are not known. This matter will be remedied in Chapter 5. Chapter 4: Transmission Line Equations 164 5. A strip transmission line of width w and separation s with s << w is the simplest example of the above summary: E = V/s n = ε dE = εdV/s q = nw C = q/V = εdw/s => K = 4πε d/C = 4π(s/w) so C = 4 πεd/K = εd (w/s) K = 4 π (s/w) G = 4 πσd/K = σd (w/s) Le = (μd/4π) K = μ d (s/w) Z0 ≈ (K /εrel ) 30Ω = 4π (s/w) (1/ εrel ) 30Ω = (s/w) (1/ εrel ) 377Ω (4.11.37) (f) Time domain equations (telegraph equations) The results above are all stated in the frequency domain , but it is a simple matter to convert them to the time domain using j ω ↔ ∂t. One then makes these replacements z = R+jωL → L ∂t + R y = G+jωC → C ∂t + G zy = (R+jω L)( G+jωC) → (R + L ∂t)( G + C∂ t) = LC∂t2 + (LG+RC) ∂t + RG . (4.11.38) Here then are selected equations and their translations to the time domain: Transmission Line Equations (4.11.14b) : [ coupled first order PDE's] ∂ zV = - z i ⇒ ∂zV(z,t) = - L ∂ti(z,t) - LR i(z,t) ∂z i = - y V ⇒ ∂z i(z,t) = - C ∂tV(z,t) - CGV(z,t) (4.11.39) Transmission Line Wave Equations (4.11.15) [ damped wave equations ] ( ∂z2 - zy) V(z) = 0 ⇒ [ ∂z2 - LC ∂t2 - (LG+RC)∂ t - RG] V(z,t) = 0 ( ∂z2 - zy) i(z) = 0 ⇒ [ ∂z2 - LC ∂t2 - (LG+RC)∂ t - RG] i(z,t) = 0 (4.11.40) If we set the loss parameters R and G both to 0 these equations become ∂zV(z,t) = -L ∂ti(z,t) [ ∂z2 - LC ∂t2] V(z,t) = 0 [ undamped wave equations] ∂z i(z,t) = -C ∂tV(z,t) [ ∂z2 - LC ∂t2] i(z,t) = 0 // lossless (4.11.41) At large ω one has L ≈ L e (note 1 above) and since (4.11.28) says L eC = μdεd = 1/vd2 we conclude that the factor LC appearing in th e above wave equations is 1/v d2 where v d is the dielectric wave velocity. The various transmission line equations shown above in the time domain are often referred to as telegraph (telegrapher, telegrapher's) equations . Chapter 4: Transmission Line Equations 165 4.12 Modifications to account for μd ≠ μ1 ≠ μ2. These modifications only affect the A z and W(z) part of this chapter, not the first six sections which are concerned with φ and V(z). So changes start with Section 4.7. If the equality μd = μ1 = μ2 assumed in Section 4.7 is broken, the result is that surface magnetization currents appear on one or both of the conductor surf aces and these cause an a lteration of the theory. Thanks to the "J m Theorem" proven in Appendix B, this alte ration can be carried through with a very minimal impact, as we now show. In Appendix B conductor magnetization surface curre nts are studied in some detail. The reader interested in how the magnetic modification is carried out would do well to read Appendix B at this point. A reader less interested can accept the Appendix B r esults and then learn below that basically nothing changes! So imagine starting with μ d = μ1 = μ2 and then changing μ1 and μ2 to new values. The question is: how do the various parameters and equations of the theory change? The first modification arises in Section 4.7. As described in Appendix B.6, the modified version of (4.7.2) is this, Az1(x) = μ1 4π ∫ C1 [ Jz1(x') + μ0 μ1 Jzm1(x') ] e-jβdR R dx'dy'dz' . R = | x - x'| (4.7.2)' where Jzm1 includes only the surface component of the magnetization current on conductor C 1. Appendix B.6 shows how this J zm1 adder term in effect adds a certain homogeneous solution to the particular solution (first term above) of the A z Helmholtz equation such that the A z boundary conditions are duly satisfied at the magnetic conductor C 1 boundary. According to (B.1.10), the surface current J zm1 when expressed in surface rather than volume notation is given by K z = - ( μ1 μ0 - μd μ0 ) Hθ and thus vanishes when μ1 = μd, resulting in the unmodified version of (4.7.2). We maintain the next two equati ons of Section 4.7 as is, involving separation of variables, J z1(x,y,z) = b 1(x,y) i1(z) A/m2 1/m2 A (4.7.3) where i 1 is scaled such that ∫C1 dx dy b1( x , y ) = 1 . (4.7.4) This i1(z) is still the total conduction current in C 1. But we now add two new equations, J z1m(x,y,z) = b 1m(x,y) i1m(z) A/m2 1/m2 A (4.12.1) Chapter 4: Transmission Line Equations 166 where i1m is scaled such that ∫C1 dx dy b1m( x , y ) = 1 . ( 4 . 1 2 . 2 ) It is understood here that b 1m(x,y) is a distribution which is restricted to the surface of C 1, but we continue to write it as if it existed at all points in the cross section of C 1. The integration in (4.12.2) is of course meant to include this surface distribution. We know from (B.1.11) and (B.1.12) that, for an arbitrarily shaped conductor C 1, i1m(z) ≡ - ( μ1 μ0 - μd μ0 ) i(z) [ μ1 = conductor C 1, μd = dielectric ], (4.12.3) and that the current ratio is therefore given by, f 1m ≡ i1m(z)/ i(z) = - ( μ1 μ0 - μd μ0 ) . ( 4 . 1 2 . 4 ) With the above definitions, our modified (4.7.6) becomes Az1(x,y,z) = μ1 4π ∫-∞ ∞ dz' i(z') ∫C1 dx' dy' [ b 1(x',y') + μ0 μ1 f1m b1m(x',y') ] e-jβdR R = μ d 4π ∫-∞ ∞ dz' i(z') ∫C1 dx' dy' [ μ1 μd b1(x',y') + μ0 μd f1m b1m(x',y') ] e-jβdR R . (4.12.5) This leads us to define a new effective transverse current density, b' 1(x,y) ≡ μ1 μd b1(x',y') + μ0 μd f1m b1m( x ' , y ' ) = μ 1 μd b1(x',y') + [1-μ1 μd ] b1m(x',y') . (4.12.6) This new transverse density b' 1 is still normalized to unity, usi ng (4.7.4) and (4.12.2) above, ∫C1 dx dy b'1(x,y) = μ1 μd ∫C1 dx dy b1(x,y) + [1-μ1 μd ] ∫C1 dx dy b'1m(x,y) = μ 1 μd * 1 + [1-μ1 μd ] * 1 = 1 . (4.12.7) Chapter 4: Transmission Line Equations 167 How does b' 1 differ from b 1? The difference is that b 1 does not include a surface current and b' 1 does. We can represent equation (4.12.6) in this symbolic graphic manner: (4.12.6) Thus, from (4.12.5) and (4.12.6) we have this new version of (4.7.6), A z1(x,y,z) = μd 4π ∫-∞ ∞ dz' i(z') ∫C1 dx' dy' b' 1(x,y) e-jβdR R . (4.7.6)' The differences are that the leading factor is μd instead of μ1, and b1 is replaced by b' 1. Moving into Section 4.8 we have this new version of (4.8.1), A z12(x) = Az1(x) + Az2(x) = μd 4π ∫-∞ ∞ dz' i(z') { ∫C1 dx1' dy1' b'1(x1',y1') e-jβdR R – ∫C2 dx2' dy2' b'2(x2',y2') e-jβdR R } ( 4 . 8 . 1 ) ' which is identical to (4.8.1) except b i → b'i. Then in the transmission line limit, we get this new version of (4.9.1), Az12(x) = μd 4π i(z) ∫-∞ ∞ dz'{ ∫C1 dx1' dy1' b'1(x1',y1') 1 R1 – ∫C2 dx2' dy2' b'2(x2',y2') 1 R2 } (4.9.1)' From this point onward, all equations are the same apart from b i → b'i. Here are some of those equations after modification: W(z) ≡ Az12(x1) - Az12(x2) ( 4 . 1 0 . 1 ) ' = μd 4π i(z) ∫-∞ ∞ dz' { ∫C1 dx1' dy1' b'1(x1',y1') 1 R11 – ∫C2 dx2' dy2' b'2(x2',y2') 1 R12 } – μd 4π i(z) ∫-∞ ∞ dz' { ∫C1 dx1' dy1' b'1(x1',y1') 1 R21 – ∫C2 dx2' dy2' b'2(x2',y2') 1 R22 } W ( z ) ( 4 . 1 0 . 3 ) ' = μd 4π i(z) ∫-∞ ∞ dz' { ∫C1 dx1' dy1' b'1(x1',y1')( 1 R11 - 1 R21 ) - ∫C2 dx2' dy2' b'2(x2',y2') (1 R12 - 1 R22 ) } . Chapter 4: Transmission Line Equations 168 W(z) = i(z) μd 4π { ∫C1 dx1' dy1' b'1(x1',y1') ln(s212/s112) - ∫C2 dx2' dy2' b'2(x2',y2') ln(s222/s122) } ( 4 . 1 0 . 4 ) ' KL ≡ ∫C1 dx1' dy1' b'1(x1',y1') ln(s212/s112) - ∫C2 dx2' dy2' b'2(x2',y2') ln(s222/s122) (4.10.9)' Le = W(z) i(z) = μd 4π KL // no change (4.10.8) We then enter Section 4.11. The derivation of the transmission line equations (4.11.14b) is unaffected by the above modifications; the only change is that the b' i appear in the integral K L in place of the b i. The derivation of the fact that K = K L ending in (4.11.30) is also unchanged! This at first seems strange since K has not changed, but we have apparently altered K L by the replacements b i → b'i. But KL is not an evaluation -- it is an integral equation relating K L to the b' i. In the self-consistent solution, the new functions (distributions) b' i adjust themselves so that K L does not change. K L cannot change because (4.11.30) says it must remain equal to K which is determ ined by the electrostatic side of the problem. It is perhaps helpful to look at (4.11.29) which says L e = μd 4π KL . We know that if the dielectric μd value does not change, the external inductance L e of the transmission line cannot change so K L stays fixed. Changing μ1 and/or μ2 away from the value μd will of course change the internal inductances of the conductors, and this is dul y noted below in terms of surface impedances. As μ1 is increased, the B field inside conductor C 1 increases (H stays the same) so the stored B field increases, and L i increases. Finally, if we look at the example associated with Fig 4.13, we still find explicitly that K L= K because the calculation leading to (4.11.33) is unchanged when b i are replaced with b' i, since the b' i are still normalized to unity as shown in (4.12.7). The happy bottom line is that all of summary box (4.11.34) is unchanged except b i → b'i in the K L integral. The constant K can still be evaluated using the "capacitor problem" of Section 5.5 below and it is unaffected by conductors having μi ≠ μd. Having said this, let us now consider what happens to an operating transmission line which starts off with μ1 = μ2 = μd = μ0 and we then gradually turn a magic "permeability knob" so that μ1 gradually increases from μ0 to some value μ1 > μ0. That is to say, we gradually cause conductor C 1 to become magnetic. The constant K (and therefore K L = K) does not change at all. This K is determined by the potential φ part of the problem in Section 4.4 and does not even know about the magnetic modification. Thus, looking at (4.11.34), C', C, G and L e do not change. In particular, L e does not change because we have not altered μd of the dielectric. The following two items shown in box (4.11.34) do change : R = Re(Z s1 + Zs2) L = L e + (1/ω) Im(Zs1 + Zs2) where Z si is the surface impedance of conductor C i. The non-L e term in L can be interpreted as the internal inductance of the conductors. R and L change because Z s1 changes if we change μ1. This is so because Z s1 is always a function of the skin depth δ1, and δ1 ≡ 2/(ωμ1σ1) from (2.2.20). In the special Chapter 4: Transmission Line Equations 169 case that C 1 is a round wire of radius a 1 with an axially symmetric current distribution (such as the center wire of a coaxial cable), we showed in (2.4 .11) that the surface impedance is given by Z1s(ω) = +jωμ1 2πa1(2/δ1) ber0[2(a1/δ1)] + j bei 0[2(a1/δ1)] ber0'[2(a1/δ1)] + j bei 0'[2(a1/δ1)] , (2.4.11) so certainly this Z s(ω) is a function of μ1 both due to the leading constant and through the five occurrences of δ1. Both the real and imaginary parts of Z 1s(ω) will change as μ1 changes, so the transmission line parameters R and L both change. In the high frequency limit , Z1s(ω) ≈ 1 σ1(2πa1)δ1 (1+j) δ << 16a , (2.4.16) so now the variation with μ1 is through the single δ1 factor shown. Again, both real and imaginary parts of Z1s(ω) vary with μ 1. Since R and L change as noted above, the transmissi on line characteristic impedance will also change, Z0 = R + jωL G + jωC = z y (K.11) This means, for example, if we drive a semi-infinite transmission line with some fixed voltage V(z), the driving current i(z) will vary in amplitude and ph ase as we turn our "permeability knob" for conductor C 1. This is simply because i(z) = V(z)/Z 0. So the good news is that the theory of Chapter 4 is easily extended to allow for magnetic conductors and or dielectric. Once again, the summary box (4.11.34) is unchanged when μ 1 = μ2 = μd is broken except for the appearance of b' i in the K L integral, and except for the fact that Z s1 and Zs2 change as noted above, causing changes in R, L and Z 0. At very high frequency, one will have Z 0 = (Le/C) and in this case Z0 is not altered, see (4.11.36). Chapter 5: The Transverse Problem 170 Chapter 5: The Transverse Problem In this Chapter we defin e a certain "transverse" potential theory problem and a prescription for the solution of that problem to obtain K and the transmission line parameters C, G and L e. 5.1 Separation of φ Let φ ≡ φ12(x) of Section 4.2. Then in the transmission line limit we found in (4.3.10) that, φ(x) = 1 4πξd q(z) ∫-∞ ∞ dz'{ ∫C1 dx1' dy1' α1(x1',y1') 1 R1 – ∫C2 dx2' dy2' α2(x2',y2') 1 R2 } . (4.3.10) Rewrite the above equation as, φ(x,y,z) = 1 4πξ d q(z) φt( x , y ) ( 5 . 1 . 1 ) φ t(x,y) = ∫-∞ ∞ dz'{ ∫C1 dx1' dy1' α1(x1',y1') 1 R1 – ∫C2 dx2' dy2' α2(x2',y2') 1 R2 } (5.1.2) where R 1 = |x-x1'|, R2 = |x-x2'|, and x is a point in the dielectric. We thus identify φt as a dimensionless "transverse potential" associated with the full potential φ. Recall that x 1 and x2 are points on the surfaces of conductors C 1 and C2 at the same z. Evaluate (5.1.1) at x1 then at x2 and then subtract to get the right equation below, V(z) = φ(x1) - φ(x2) = 1 4πξd q(z) [φt(x1,y1) - φt(x2,y2)] . The left side is just V(z) according to (4.4.1). Recalling now from (4.4.7) that V(z) = q(z) 1 4πξ d K (4.4.7) we conclude that φ t(x1,y1) - φt(x2,y2) = K . ( 5 . 1 . 3 ) The Helmholtz equation for φ is given by (1.5.3) for a region including dielectric and conductors, ( ∇2 + βd2)φ(x,y,z) = - (1/ εd) ρ(x,y,z) (1.5.3) (5.1.4) where ρ(x,y,z) exists on the boundary of the dielectric region (ie, on the conductor surfaces). Inside the dielectric there is no ρ so we then have Chapter 5: The Transverse Problem 171 ( ∇2 + βd2)φ(x,y,z) = 0 . // dielectric region (5.1.5) Inserting (5.1.1) into (5.1.5) yields, ( ∇ 2 + βd2) 1 4πξd q(z) φt(x,y) = 0 or ( ∇ t2 + ∂z2 + βd2) 1 4πξd q(z) φt(x,y) = 0 // ∇t2 = ∇2D2 = ∇2 - ∂z2 or ∇t2φt(x,y) q(z) + φt(x,y) ∂z2q(z) + βd2 φt(x,y) q(z) = 0 . Divide through by φt(x,y) q(z) to get ∇t2φt(x,y) φt(x,y) + ∂z2 q(z) q(z) + βd2 = 0 or [ ∇ t2φt(x,y) φt(x,y) ] + ∂z2 q(z) q(z) = - βd2 (5.1.6) which has the general form, [ h(x,y) ] + g(z) = - βd2 . The only way this can be true for all x,y,z in a region is if g(z) = some constant, which call - k φ2. Then, ∂z2 q(z) q(z) = - k φ2 ∇t2φt(x,y) φt(x,y) = - βd2 + kφ2 . (5.1.7) We can rewrite these equations as [ ∇t2 + (βd2 - kφ2)] φt(x,y) = 0 (5.1.8) [ ∂z2 + kφ2] q(z) = 0 . (5.1.9) According to Fact (3.8.10) and (5.1.1), for a particular z value, we expect φ t(x,y) to have some constant value K1 on the entire perimeter of a cross section of conductor C 1, and some other constant value K 2 on the entire perimeter of a cross section of conductor C 2, These facts act as boundary conditions for (5.1.7). φ t(C1) = K1 φt(C2) = K2 K 1 - K2 = K (5.1.10) so that (5.1.3) is realized. The second equation (5.1.9) has the following solution Chapter 5: The Transverse Problem 172 q(z) = q(0) e-jkφz => q(z,t) = q(0) ej(ωt-kφz) (5.1.11) and we find that q(z) has the form of a wave tr aveling down the transmission line with wavenumber k φ. The reader of Chapter 2 or of Appendix D will recognize this as the form assumed for the electric field in (2.1.1) or (D.1.1) where it was a ssumed as an ansatz without much a priori justification. For example, E(r,θ,z,t) = e j(ωt-βdz) E(r,θ) . (D.1.1) When the dust settles below, for a low-loss transm ission line we shall in fact end up with k φ = βd so that (D.1.1) has the same traveling wave form as (5.1.11). 5.2 Separation of A z Let Az ≡ Az12(x) of Section 4.8. Then in the transmission line limit we found in (4.9.1) that Az(x) = μd 4π i(z) ∫-∞ ∞ dz'{ ∫C1 dx1' dy1' b1(x1',y1') 1 R1 – ∫C2 dx2' dy2' b2(x2',y2') 1 R2 } . (4.9.1) Rewrite the above equation as, Az(x,y,z) = μd 4π i(z) Azt( x , y ) ( 5 . 2 . 1 ) Azt(x,y) = ∫-∞ ∞ dz'{ ∫C1 dx1' dy1' b1(x1',y1') 1 R1 – ∫C2 dx2' dy2' b2(x2',y2') 1 R2 } (5.2.2) where R 1 = |x-x1'|, R2 = |x-x2'|, and x is a point in the dielectric. We thus identify A zt as a dimensionless "transverse vector potential" associated with the full vector potential A z. Recall that x 1 and x2 are points on the surfaces of conductors C 1 and C2 at the same z. Evaluate (5.2.1) at x1 then at x2 and then subtract to get the right equation below, W(z) = A z(x1) - Az(x2) = μd 4π i(z) [A zt(x1,y1) - Azt(x2,y2)] . The left side is just W(z) according to (4.10.1). Recalling now L e = W(z) i(z) = μd 4π KL => W(z) = μd 4π i(z) KL (4.10.8) we conclude that A zt(x1,y1) - Azt(x2,y2) = KL. Chapter 5: The Transverse Problem 173 But (4.11.30) says K L = K, so write this last as Azt(x1t) - Azt(x2t) = K . (5.2.3) The Helmholtz equation for A z is given by (1.5.4) for a region including dielectric and conductors, ( ∇2 + βd2)Az(x,y,z) = - Σi=2N μiJi . (1.5.4) (5.2.4) These Ji are currents inside the conductors. Although there is small conduction current in the dielectric, it has been absorbed into β2 as shown in (1.3.21) in the time domain with the use of the King gauge. If we take our region of interest to be the dielectric alone, we then have ( ∇2 + βd2)Az(x,y,z) = 0 . // dielectric region (5.2.5) Inserting (5.2.1) into (5.2.5) yields, ( ∇2 + βd2) μd 4π i(z) Azt(x,y) = 0 or ( ∇t2 + ∂z2 + βd2) μd 4π i(z) Azt(x,y) = 0 or ∇t2Azt(x,y) i(z) + A zt(x,y)∂z2i(z) + βd2 Azt(x,y) i(z) = 0 . Now divide through by A zt(x,y) i(z) to get [∇t2 Azt(x,y) Azt(x,y) ] + ∂z2 i(z) i(z) + βd2 = 0 (5.2.6) which has the general form, [ h(x,y) ] + g(z) = - β d2 . The only way this can be true for all x,y,z in a region is if g(z) = some constant, which call -k A2. Then, ∂z2 i(z) i(z) = - kA2 ∇t2 Azt(x,y) Azt(x,y) = - βd2 + kA2 . (5.2.7) We can rewrite these equations as [ ∇t2 + (βd2 - kA2)] Azt(x,y) = 0 (5.2.8) [ ∂z2 + kA2] i ( z ) = 0 . ( 5 . 2 . 9 ) Chapter 5: The Transverse Problem 174 According to Fact (3.8.11) and (5.2.1), for a particular z value, we expect Azt(x,y) to have some constant value W 1 on the entire perimeter of a cross section of conductor C 1, and some other constant value W 2 on the entire perimeter of a cross section of conductor C 2, These facts act as boundary conditions for (5.1.7). Since a potential has an arbitrary zero, we shall set A zt(C1) = W1 Azt(C2) = W2 W 1 - W2 = K (5.2.10) so that (5.2.3) is realized. The second equation (5.2.9) has the following solution i(z) = i(0) e -jkAz => i(z,t) = i(0) ej(ωt-kAz) (5.2.11) and we find that i(z) has the form of a wave tr aveling down the transmission line with wavenumber k A. Comparing (5.2.11) with (5.1.11), it would certainly seem odd if q(z) and i(z) had the form of traveling waves with different wavenumbers k φ ≠ kA. We will formally show in the next section that k φ = kA. 5.3 Development of the Transverse Problem (a) kφ = kA and the transverse equations The longitudinal equations from the previous two sections are these: [ ∂ z2 + kφ2 ] q ( z ) = 0 (5.1.8) [ ∂z2 + kA2 ] i ( z ) = 0 . (5.2.8) But, φ(x,y,z) = 1 4πξd q(z) φt( x , y ) (5.1.1) Az(x,y,z) = μd 4π i(z) Azt( x , y ) . (5.2.1) Therefore, [ ∂ z2 + kφ2 ] φ(x,y,z) = 0 [ ∂z2 + kA2 ] Az(x,y,z) = 0 . (5.3.1) Recall that x 1 and x2 are points on the surfaces of conductors C 1 and C2. If we write equations (5.3.1) first at x1 and then at x 2 and then subtract, we get longitudinal equations for V(z) and W(z), [ ∂ z2 + kφ2 ] V(z) = 0 // V(z) = φ(x1) - φ(x2) [ ∂z2 + kA2 ] W(z) = 0 // W(z) = A z(x1) - Az(x2) (5.3.2) Chapter 5: The Transverse Problem 175 where we have used the definitions V(z) and W(z) from (4.4.1) and (4.10.1). For low frequencies, we average (5.3.1) over the conductor perimeters and then V(z) and W(z) are as in (4.11.7). Recall now the second order transmission line equations of (4.11.15), d 2V(z) dz2 - zy V(z) = 0 d2i(z) dz2 - zy i(z) = 0 . (4.11.15) (5.3.3) Comparison of (5.3.2) with (5.3.3) shows that kφ2 = kA2 ≡ k2 = -zy = - (R+j ωL)(G+jωC ) ( 5 . 3 . 4 ) which fulfills the expectation earlier that we should have k φ = kA. With the longitudinal behavior, q(z) = q(0) e-jkφz => q(z,t) = q(0) ej(ωt-kφz) (5.1.11) i(z) = i(0) e-jkAz => i(z,t) = i(0) ej(ωt-kAz) (5.2.11) the appropriate root for k is then k = -j zy => jk = zy ≡ (R+jωL)(G+jωC) = a + jb . // a and b are real and imag parts of jk (5.3.5) Then all quantities like q(z), i(z),V(z),W(z) have this longitudinal behavior for a wave traveling in the +z direction, F(z) = F(0) e -jkz = F(0) e-az e-jbz jk = a + jb = zy = (R+jωL)(G+jωC) = F(0) exp[ - zy z] = F(0) exp[ - (R+jωL)(G+jωC) z] a ≡ Re( zy ) = Re[ (R+jωL)(G+jωC) ] = - Im(k) // attenuation per distance of F(z) b = Im( zy ) = Im[ (R+jωL)(G+jωC) ] = Re(k) . // phase of F(z) (5.3.6) Now recall from box (4.11.34) that z = Zs + jω Le = Zs + jω μd 4π K Z s ≡ Zs1 + Zs2 y = jω C' = jω 4πξd/ K ( 5 . 3 . 7 ) so k2 = -zy = -[Zs + jωμd 4π K] jω 4πξd/K = - Z s jω 4πξd/K + ω2μdξd = - j ω Z s 4πξd/K + βd2 . // see (1.5.1a) (5.3.8) Chapter 5: The Transverse Problem 176 Therefore (βd2 - k2) = jω Zs 4πξd / K = jω Zs 4πξd K = jω Zs C' = j ω Zs(ξd/εd) C = jω (1/εd) [εd + σd/jω] ZsC = [jω + σd/εd] ZsC (5.3.9) The transverse equations (5.1.8) and (5.2.8) and boundary conditions (5.1.10) and (5.2.10) may now be summarized: [ ∇ t2 + (βd2-k2)] φt(x,y) = 0 φt(C1) = K1 φt(C2) = K2 K1- K2 = K (5.3.10) [ ∇t2 + (βd2-k2)] Azt(x,y) = 0 A zt(C1) = W1 Azt(C2) = W2 W1- W2 = K (5.3.11) where ( βd2- k2) = jω Zs 4πξd K . (b) The scaling boundary condition on φt(x) There exists another boundary condition on φt in the case that the dielect ric extends transversely to infinity. Recall (5.1.2) for φ t(x,y) = φt(x), φt(x) = ∫-∞ ∞ dz'{ ∫C1 dx1' dy1' α1(x1',y1') 1 R1 – ∫C2 dx2' dy2' α2(x2',y2') 1 R2 } (5.1.2) R12 = (x-x1')2 + (y-y1')2 + (z-z')2 = s12 + (z-z')2 s 12 = (x-x 1')2 + (y-y1')2 (4.2.1) R22 = (x-x2')2 + (y-y2')2 + (z-z')2 = s22 + (z-z')2 s22 = (x-x 2')2 + (y-y2') . If we take the point x transversely far away from the conduc tors, the following drawing shows the distances R 1 and R2 which appear in the above integration, Chapter 5: The Transverse Problem 177 Fig 5.1 During the transverse integration ∫C1 dx1' dy1', distance R 1 does not vary much and can be replaced with a distance from x to the "center" of conductor C 1 without changing the integral significantly. The same can be said for R 2. We shall refer to these "center points" as x1 and x2 (this is a new and different use for these variable names). In this case, we obtain φt(x) ≈ ∫-∞ ∞ dz' { 1 R1 ∫C1 dx1' dy1' α1(x1',y1') – 1 R2 ∫C2 dx2' dy2' α2(x2',y2') } = ∫-∞ ∞ dz' { 1 R1 - 1 R2 } = ∫-∞ ∞ dz' (1 s12 + (z-z')2 - 1 s22 + (z-z')2 ) (5.3.12) where we have used the fact (4.1 .3) that the transverse charge densities are normalized to unity. The dz' integral was done in (4.4.5) and equals ln(s 22/s12), so then φt(x) ≈ ln(s22/s12) // limiting form as point x = (x,y) moves far from the conductors ( 5 . 3 . 1 3 ) s12 = (x-x1)2 + (y-y1)2 s22 = (x-x2)2 + (y-y2)2 . Whatever the exact solution φt(x) might be, in the limit discussed above one must obtain φt(x) ≈ ln(s22/s12). Of course as one continues to move x away to infinity, s 1 ≈ s2 and then φ t(x) ≈ ln(1) = 0. Basically (5.3.13) is a boundary cond ition on the "scale" of the solution φt(x). If someone were to propose a possible solution φt(x) = 2.6 ln(s 22/s12) for some conductor geometry, we could instantly rule Chapter 5: The Transverse Problem 178 out that solution since it violates the b oundary condition (5.3.13). The scale of φt is restricted in this manner because the charge distributions α i appearing in (5.3.12) are normalized to unity. By the exact same argument presented above, we have A zt(x) ≈ ln(s22/s12) // limiting form as point x = (x,y) moves far from the conductors (5.3.14) We shall give an interpretation of these limiting forms in Section 5.4 (b) below. (c) Energy Conservation in a Transmission Line In (5.3.6) we have seen how the voltage or cu rrent in a transmission line has z dependence e-jkz where k = -j zy . (5.3.6) Now consider the following quantities: u C = (1/2) C V(z)2dz = capacitative energy stored in dz uL = (1/2) L i(z)2dz = inductive energy stored in dz (as in (C.3.5)) p C = Cdz V(z) ∂ tV(z) = rate of increase of the C stored energy pL = Ldz i(z) ∂ t i(z) = rate of increase of the L stored energy p R = i(z)2Rdz = rate of energy burned in R pG = V(z)2Gdz = rate of energy burned in G // V(z) i G(z) = V(z) [ V(z) Gdz ] p(z) = energy/sec entering a little transmission line segment of length dz located at z p(z+dz) = energy/sec leaving the segment at z + dz (5.3.15) The power balance equation for the transmission line segment of length dz is then p(z)-p(z+dz) = power flow decrease over dz = p C + pL + pR + pG = Cdz V(z) ∂tV(z) + Ldz i(z) ∂ t i(z) + i(z)2Rdz + V(z)2Gdz (5.3.16) so that - ∂ zp(z) = CV(z) ∂ tV(z) + L i(z) ∂t i(z) + i(z)2R+ V(z)2G or - ∂ zp(z) = ∂t [ (1/2)CV(z)2 + (1/2)Li(z)2 ] + i(z)2R+ V(z)2G Since p(z) = V(z)i(z), one finds - ∂ z[V(z,t) i(z,t)] = ∂t[ (1/2)CV(z,t)2 + (1/2)Li(z,t)2 ] + i(z,t)2R+ V(z,t)2G (5.3.17) Chapter 5: The Transverse Problem 179 where we now show both space and time arguments. One can regard the above as a statement of energy conservation (per unit time) at location z on an infini te transmission line. For a lossless line, R = G = 0, and it is for such a line that the above equation appears in Haus and Melcher as Sec 14.2 Eq. (19). Verification check Moving the time derivative back in one gets, - ∂ z[V(z,t) i(z,t)] = [ CV(z,t) ∂ tV(z,t) + L i(z,t) ∂t i(z,t) ] + i(z,t)2R+ V(z,t)2G . Since both V and i have the z dependence e -jkz, V i has dependence e-2jkz so, -2jk [V(z,t) i(z,t)] = [ CV(z,t) ∂ tV(z,t) + L i(z,t) ∂t i(z,t) ] + i(z,t)2R+ V(z,t)2G . Taking ∂t → jω and writing V(z, ω) = V and i(z, ω) = I, the above becomes in the ω domain, -2jk [VI] = j ω[ CV2 + LI2 ] + I2R+ V2G = I 2 (R + jωL) + V2(G + jωC) = I2 z + V2 y . Dividing both sides by VI gives -2jk = (I/V) z + (V/I) y . But (4.11.16) says that V/I = Z 0 = z/y so we find, -2jk = (I/V) z + (V/I) y = y/z z + z/y y = zy + zy = 2 zy and we finally arrive at -jk = zy which matches the equation stated at the start of this subsection. 5.4 The Low-Loss Approximation (a) Transverse Equations for a Low-Loss transmission line For low loss, we take the conductor surface impedance Z s ≈ 0. Recall from (5.3.8) that k 2 = βd2 - jω Zs 4πξd/ K . (5.3.8) Our definition of a "low-loss" transmission line is one for which k 2 ≈ βd2 and in this case the longitudinal wave number k as shown in (5.1.11) and (5.2.11) is k = β d. So our low-loss condition is (using (1.5.1a) for βd2), Chapter 5: The Transverse Problem 180 | jω Zs 4πξd/K| << |βd2| β d2 = ω2μd ξd o r |Z s| << (1/4 π) | βd2/(ωξd)| K = (1/4 π) ω | βd2/(ω2ξd)| K = (1/4π ) ωμd K so |Zs| << (1/4 π) ωμd K . ( 5 . 4 . 1 ) For a symmetric-environment round wire of radius a we found in (2.4.12) that for large ω , Z s ≈ 1 σ(2πa)δ (1+j) for δ << 16a δ2 = 2/ωμσ . (2.4.16) For a transmission line of two round conductors e ither coaxial or widely spaced we can estimate Zs = Zs1 + Zs2 = 1 σ(2π)δ (1+j) ( 1 a1 + 1 a2 ) ≡ 1 σ(2π)δ (1+j) 1 a 1 a ≡ ( 1 a1 + 1 a2 ) so that (5.4.1) says [assuming μ = μd ] σ | Zs | = 1 (2πa)δ 2 << (1/4 π) σωμ K = (1/4 π) (2/δ2) K => 1 (2πa)δ 2 << (1/4 π) (2/δ2) K => 1 δa 2 << (1/δ2) K => (δ/a) << K/ 2 ( 5 . 4 . 2 ) We saw in the Example of Section 4.6 that K = 2 ln(a 2/a1) for a coaxial cable. Even for a very large radius ratio of 100 this would be K = 2 ln(100) = 9.2. For a more typical ratio of perhaps 5, K ≈ 3.2. Then our inequality above says roughly (δ/a) << 2 1 a ≡ ( 1 a1 + 1 a2 ) which is then our ball-park estimate for applicability of the "low-loss transmission line" condition at large ω. We showed in Section 2.5 (and Section 4.11) how Z s can be modified for some other geometry. Basically this says we are in the low-loss limit if the sk in depth is much smaller than the wire's transverse dimensions. On the other hand, for small ω we found in (2.4.12) that Zs(ω) = Rdc + jω μ 8π = Rs + jωLs // low frequency limit (2.4.12) Chapter 5: The Transverse Problem 181 where R dc = 1/(σπa2) for a round wire. If we use this as an estimate for Z s of each conductor in the case of general conductors, then Zs = Zs1 + Zs2 = Rdc1 + Rdc2 + 2jω μ 8π ≡ RDC + 2jω μ 8π and then (5.4.1) says | Zs| = | RDC + 2jω μ 8π | << (1/4 π) ωμ K or (RDC)2 + (μ/4π)2ω2 << (μ/4π)2ω2 K2 RDC << (μ/4π) ω K2-1 . For a given low frequency ω, RDC must be smaller than the above for the transmission line to be low-loss. Low ω Example 1: Belden 8281 coaxial cable is treated as a case study in Appendix R. There it is shown that R DC = .036 ohm/m and K = 3.7. The inequality above then requires that ω >> (4π/μ) RDC 1/K2-1 = 107 * .036 / 3.56 ≈ 105 => f >> 16 KHz So in the low frequency range, as long as f is not too low, one can treat 8281 cable as low-loss. Low ω Example 2: At the end of Section 4.5 we consider ed a power transmission line with two 1" diameter conductors separated by 1 meter. It was found that K = 17.5 and that R dc = .02Ω per thousand feet for each conductor which is 0.66 x 10-4 ohms/m for each conductor. Thus we need ω >> (4π/μ) RDC 1/K2-1 = 107 * [2* 0.66 x 10-4] / 17.47 ≈ 76 => f >> 12 Hz Such power lines are normally operated at 50 or 60Hz so are in the low loss regime. In any event, if we assume this low-loss limit is in effect, then β d2- k2 = jω Zs 4πξd/K ≈ 0 and our transverse equations (5.3.10) and (5.3.11) become 2D Laplace equations, ∇t2φt(x,y) = 0 φt(C1) = K1 φt(C2) = K2 K1- K2 = K (5.4.3) ∇t2Azt(x,y) = 0 A zt(C1) = W1 Azt(C2) = W2 W1- W2 = K (5.4.4) As commented earlier, the parallelism between φt and Azt should not be surprising in light of Section 1.3 (b) where it was noted that A and φ are components of the same relativistic 4-vector. Chapter 5: The Transverse Problem 182 (b) The scaling boundary condition (5.3.13) revisited First, a quick review. In 3D the potential (SI units) of a point charge q located at x1 is φ(x) = (q/4πε|x-x1|) = q/(4πε R1). In 2D the potential of a point charge q located at x1 is φ(x) = -(q/2πε) ln|x-x1| = -(q/2πε) lns1. The 3D φ(x) is the solution of - ∇2(φ) = (q/ε) δ3)(x-x1) as shown in (H.1.4) and as proven in Appendix H. The quantity 1/4 πR1 is the 3D free-space propagator of the 3D Laplace equation. It is the Green's function of the equation - ∇2g(x|x1) = δ(3)(x-x1) . The 2D φ(x) is the solution of -∇2 2D(φ) = (q/ε) δ(2)(x-x1) as shown in (I.1.4) and as proven in Appendix I. The quantity -2 πlns1 is the 2D free-space propagator of the 2D Laplace equation. It is the Green's function of the equation - ∇2 2Dg(x|x1) = δ(2)(x-x1). With this brief review, we now examine a 2D cross section view of the transmission line of Fig 5.1 at a scale that makes the two conductors appear very small and very close together, and at the same time we imagine more elaborate cross section shapes. The dielectric is assumed non-conducting, so ξd = εd. The three points indicated by the three dots on the right all lie in the plane of paper; this is just a 2D drawing and for example x = (x,y). F i g 5 . 2 The dots on the left indicate the "center of charge" for each conductor and these dots appear also on the right. The claim is that when x is very far away, the variation in s 1 as it moves over the perimeter of the conductor C 1 cross section is so small that we can replace s 1 with a distance to the center of charge of C 1 and similarly for R 2. Thus, on the right we end up with the 2D potential of two point charges which form a 2D electric dipole. Using the results just quoted in the above review, we find that φ(x) = φ1(x) + φ2(x) = -(q/2πεd)ln|x-x1| -(-q/2πεd)ln|x-x2| = -(q/2 πεd) lns1+(q/2πεd) lns2 = ( q / 2 πε d) ln(s2/s1) = (q/4πεd)ln(s22/s12) . Chapter 5: The Transverse Problem 183 Recalling for ξd = εd that φ(x,y,z) = 1 4πεd q(z) φt( x , y ) (5.1.1) we find that φ t(x,y) = ln(s 22/s12) . // for r far away Thus we have an alternate derivation and 2D dipole interpretation of our earlier "scaling boundary condition" (5.3.13). 5.5 The Capacitor Problem We have now boiled down the computation of transmi ssion line parameters (in the transmission line limit and in the low-loss limit) to the problem of compu ting the capacitance of a section of transmission line. Here we assume the dielectric is non-conducting so ξ = ε and we don't have to worry about the distinction between charge densities q c and qs as discussed in (4.11.23). Solving the capacitor problem using φ A standard approach to a general 2D electrostatics capacitor problem is as follows. Start with ∇ 2D2φ(x,y) = 0 φ(C1) - φ(C2) = V = voltage between conductors (5.5.1) where we now (arbitrarily) use notation ∇ 2D2 in place of ∇t2. Since the dielectric presumably fills the region be tween the conductors, the dielectric is the official "region" of a Green's function problem. If we put a unit positive point charge at some location (x',y') in the dielectric region we can then formally (!) solve this 2D Green's function problem, − ∇2D2g(x,y|x',y') = δ(x-x')δ(y-y') g(x,y|x',y') = 0 for (x,y) on both C 1 and C2 g(x,y|x',y') = 0 for (x,y) = ∞ (if appropriate) (5.5.2) Here g(x,y|x',y') is specific to our geometry; it is not the 2D free-space Green's function - ln(1/R)/2 π shown in (I.1.4). The free-space solution has only th e lower boundary condition stated above. We assume now that this Green's function problem has been solved, either analytically, approximately, or numerically, so that g(x,y|x',y') is known (example coming in Chapter 6). In very general notation, if a region contains some sources q(x) and if the potential φ is prescribed on the entire closed boundary surrounding the region by a function f, then the solution to (5.5.1) is given in Stakgold notation as (1.5.11) (which we derive in the lines following (1.5.11) for both Laplace and Helmholtz equations ), φ(x) = ∫R dx' g(x|x') q(x') – ∫σ dSξ f(ξ) ∂ξng(x|ξ) // Stakgold (6.81) . (1.5.11) Chapter 5: The Transverse Problem 184 where σ represents the closed boundary of the region of interest, dS ξ is an integration over this boundary, and ∂ξng(x|ξ) is the derivative of the Green's function in a direction locally normal to the boundary surface. In potential theory, this type of problem is known as "the Dirichlet Problem". In our case the boundary consists of C 1, C2 and the Great Circle at ∞. Stakgold deals in an arbitrary number of spatial dimensions, but we have only 2 dimensions here, so dS ξ is a line integral around the boundary. A picture is in order, showing a cross section of the transmission line, Fig 5.3 We know that on the great circle φ = 0, so there will be no contribution from that part of the Dirichlet boundary. What we do not know are V 1 and V2 which are the constant potentials on C 1 and C2. If it happened that the picture had mirror symmetry in a plane separating the two conductors, we would know that V1 = V/2 and V 2 = -V/2, but in the general case we don't know V 1 and V2 a priori. For the moment, we leave them as to-be-determined quantities. In our application of (1.5.11) there are no charges q(x) in the dielectric region. We put one there temporarily to obtain the Green's function, but it is now gone. Thus (1.5.11) reads φ(x,y) = – ∫{C1 ds' f(C1) ∂ng(x,y|x',y') – ∫{C2 ds' f(C2) ∂ng(x,y|x',y') - ∫{ GC ds' f(∞) ∂ng(x,y|x',y') = – ∫{C1 ds' V1 ∂ng(x,y|x',y') – ∫{C2 ds' V2 ∂ng(x,y|x',y') – ∫{ GC ds' (0) ∂ng(x,y|x',y') = - V 1 ∫{C1 ds' ∂ng(x,y|x',y') – V 2 ∫{C2 ds' ∂ng(x,y|x',y') }. Chapter 5: The Transverse Problem 185 = V 1 F1(x,y) + V 2F2( x , y ) ( 5 . 5 . 3 ) where the F i(x,y) are determined by doing the line integrals for a given geometry. If y = y 1(x) describes a piece of the C 1 perimeter, then ds' = dx'2 + dy'2 = 1 + ∂x'y1(x') dx' (5.5.4) which gives a candidate ds' for doing the line in tegral over that piece of the perimeter. Once φ(x,y) is known, one can compute the normal electric field E n at the conductor surfaces, En(x) = - ∂n1φ(x) = -V1 [∂n1F1(x)] - V2 [∂n1F2(x)] ≡ V1 G11(x) + V2 G12(x) x on C1 En(x) = - ∂n2φ(x) = -V1 [∂n2F1(x)] - V2 [∂n2F2(x)] ≡ V1 G21(x) + V2 G22(x) . x on C2 (5.5.5) Since the conductors are different, the resulting four functions G ij will in general be different. For example, we are taking normal derivatives of the F i at different points in space on different (1D) surfaces. Fig 5.3 is meant to represent the 2D cross secti on of a 3D transmission line, and in the following the symbol n refers to the true surface charge density in Cou/m2. We compute n using (1.1.47) assuming E n = 0 inside the conductor, n1(x,y) = εdEn(x,y) = εdV1 G11(x) + εdV2 G12(x) x on C1 n2(x,y) = εdEn(x,y) = εdV1 G21(x) + εdV2 G22(x) x on C2 (5.5.6) where ε d is of course for the dielectric. One can then integrate over the boundaries of the conductors to get the total charges q 1 and q2 residing on the conductors (per unit length), q1 = ∫{C1 ds' n1(x',y') = ε dV1H11 + εdV2H22 q2 = ∫{C2 ds' n2(x',y') = εdV1H21 + εdV2H22 ( 5 . 5 . 7 ) where the H ij are now four constants which we have computed by doing the above process. Since it turns out that H 12 = H21 as shown below, we can ignore H 21, G21(x), and ∂n2F1(x) in the above set of calculations. We now define some new constants c ij = εdHij and write the above as q 1 = c11V1 + c12V2 q2 = c21V1 + c22V2 . (5.5.8) Comment : The coefficients c ij are dimensionally capacitance, but they are a little strange. If we start off with the conductors holding charges q 1 and q'2 and then we ground C 2 to the great circle (thin wire, V 2= 0), and then we measure V 1 relative to the great circle, we find that V 1 = q1/c11 and q2 = c21V1. So c11 is the capacitance of C 1 in the presence of a grounded C 2 (which is not the same as the capacitance of C 1 in Chapter 5: The Transverse Problem 186 isolation). And c 21 determines how much charge q 2 is "induced" onto C 2 by the presence of charged C 1. Smythe (p 37) and Oughstun (p 23) refer to the c ij both as "coefficients of capacitance" and "coefficients of induction". This should be distin guished from the notion of conductors C 1 and C2 each having a "self- capacitance" (each in isolation) and having a "m utual capacitance" ( to be called C below). Writing the above pair of equations in matrix notation we get, ⎝⎛ ⎠⎞ q1 q2 = ⎝⎛ ⎠⎞ c11 c12 c21 c22 ⎝⎛ ⎠⎞ V1 V2 or q = c V . (5.5.9) We know all the c ij because we computed them above. Then invert to get ⎝⎛ ⎠⎞ V1 V2 = ⎝⎛ ⎠⎞ s11 s12 s21 s22 ⎝⎛ ⎠⎞ q1 q2 or V = sq (5.5.10) where matrix s = c-1 is called the "mutual elastance" matrix by Smythe (p 36), and the "coefficients of potential" by Oughstun (p 21). Both authors d eal with an arbitrary number of conductors. The reader will not be surprised to learn that in general c ij = cji and sij = sji so the matrices c and s are in fact symmetric matrices. Smythe shows th is on pages 36-37 based on what he calls "Green's Reciprocation Theorem" on page 34 (George Green once again!). This theorem can be a lifesaver in certain electrostatic problems. Now our problem as shown in Fig 5.3 is to compute the potential φ when C 1 has charge q and C 2 has charge -q. We then finally arrive at the appropriate values of V 1 and V2 for our problem, which we said above were "to be determined". Here they are: ⎝⎛ ⎠⎞ V1 V2 = ⎝⎛ ⎠⎞ s11 s12 s21 s22 ⎝⎛ ⎠⎞ q -q = q ⎝⎛ ⎠⎞ s11 s12 s21 s22 ⎝⎛ ⎠⎞ 1 -1 ( 5 . 5 . 1 1 ) so that V 1 = q (s11- s12) V2 = q (s21- s22) V = V 1 - V2 = q [s11+ s22 - 2s12] . // s 12 = s21 as noted above (5.5.12) Finally, we have computed the (inverse) capacitance of our transmission line section, 1/C = V/q = s 11 + s22 - 2s12 . But we know how to invert a simple 2x2 matrix (T = transpose, cof = cofactor, det(c) = |c| ) s = c-1 = cof(cT)/det(c) so that Chapter 5: The Transverse Problem 187 s = ⎝⎛ ⎠⎞ s11 s12 s21 s22 = ⎝⎛ ⎠⎞ c22 -c12 -c21 c11 /det(c) . (5.5.13) Then 1/C = s 11 + s22 - 2s12 = ( c22 + c11 +2c12)/det(c) = c11+ c22 + 2c12 c11c22 - c122 so C = c11c22 - c122 c11+ c22 + 2c12 . (5.5.14) We have found verification of this resu lt on the web from Oughstun page 27, Once we have C, we know from (4.11.34) that K = 4 πε d/C = 4πεd c11+ c22 + 2c12 c11c22 - c122 . (5.5.15) Thus, we have solved "the capacitor problem" to obtain K for the transmission line. The other line parameters are then given as in (4.11.34) G = 4 πσ d/K L e = μd 4π K . Statement of the capacitor problem in terms of φ t To show that our capacitor problem is the same as (5 .4.3), we first quote the capacitor problem (5.5.1), ∇2D2φ(x,y) = 0 φ(C1) - φ(C2) = V . (5.5.1) Then use (5.1.1) that φ(x,y,z) = q(z) 4πε φt(x,y) to get ∇ 2D2φt(x,y) = 0 q(z) 4πε φt (C1) - q(z) 4πε φt(C2) = V or ∇ 2D2φt(x,y) = 0 φt (C1) - φt(C2) = V 4πε q(z) or ∇2D2φt(x,y) = 0 φt(C1) - φt(C2) = K Chapter 5: The Transverse Problem 188 which is (5.4.3). In the last st ep we used (4.4.7) that V(z) = q(z) 4πε K. The potentials V 1 and V2 are related to constants K 1 and K2 by V 1 = q(z) 4πε K1 V2 = q(z) 4πε K2 ( 5 . 5 . 1 6 ) Solution of the capacitor problem using φ t Here we just repeat the above analysis, showing ho w things differ. We leave out the words. The main differences are that the V i are replaced by K i and the factor q(z) 4πε appears on the lines where n i are computed. As before, we now start off with K 1 and K2 unknown, but we find them in the end: φt(x,y) = – ∫{C1 ds' K1 ∂ng(x,y|x',y') – ∫{C2 ds' K2 ∂ng(x,y|x',y') = K 1 F1(x,y) + K 2F2(x,y) En(x,y) = - ∂n1φ = - q(z) 4πε ∂n1φt(x,y) = q(z) 4πε { K1 G11(x) + K2 G12(x) } x on C1 En(x,y) = - ∂n1φ = - q(z) 4πε ∂n2φt(x,y) = q(z) 4πε {K1 G21(x) + K2 G22(x) } x on C2 n 1(x,y) = εEn(x,y) = q(z) 4πε εK1 G11(x) + q(z) 4πε εK2 G12(x) x on C1 n2(x,y) = εEn(x,y) = q(z) 4πε εK1 G21(x) + q(z) 4πε εK2 G22(x) x on C2 q1 = ∫{C1 ds' n1(x',y') = q(z) 4πεd [εdK1H11 + εdK2H22] = q(z) 4πε [ c11V1 + c12V2 ] q2 = ∫{C2 ds' n2(x',y') = q(z) 4πεd [εdK1H21 + εdK2H22] = q(z) 4πε [ c21V1 + c22V2 ] ⎝⎛ ⎠⎞ q1 q2 = q(z) 4πεd ⎝⎛ ⎠⎞ c11 c12 c21 c22 ⎝⎛ ⎠⎞ K1 K2 or q = q(z) 4πεd c K . ⎝⎛ ⎠⎞ K1 K2 = 4πεd q(z) ⎝⎛ ⎠⎞ s11 s12 s21 s22 ⎝⎛ ⎠⎞ q1 q2 or K = 4πεd q(z) s q ⎝⎛ ⎠⎞ K1 K2 = 4πεd q(z) ⎝⎛ ⎠⎞ s11 s12 s21 s22 ⎝⎛ ⎠⎞ q(z) -q(z) = 4πεd ⎝⎛ ⎠⎞ s11 s12 s21 s22 ⎝⎛ ⎠⎞ 1 -1 K1 = 4πεd (s11- s12) Chapter 5: The Transverse Problem 189 K2 = 4πεd (s21- s22) K = K 1 - K2 = 4πεd [s11+ s22 - 2s12] so K = 4πεd c11+ c22 + 2c12 c11c22 - c122 ( 5 . 5 . 1 7 ) Then the same capacitance shown in (5.5.14) is recovered, C = 4 πε d/K = c11c22 - c122 c11+ c22 + 2c12 . For arbitrary conductor shapes, carrying out the Green' s function program just outlined is quite difficult and usually requires expanding the Green's function in some complete set of eigenfunctions and then making various approximations. Perhap s conformal mapping is helpful in certain cases. Our point is that the capacitor problem is a well-posed problem and has a solution value K. Numerical evaluations are always possible as noted earlier. If the conductors are round, the problem can be solved exactly as we shall show in Chapter 6. 5.6 What happens if low-loss is not assumed? We have s een how one can analyze a transmission line in the low-loss regime by studying the associated capacitor problem. The reader is reminded that the term low-loss does not mean no-loss! A low-loss transmission line does have losses, meaning it has atte nuation. This attenuation is associated with the imaginary part of k as shown in (5.3.6) and as examin ed in Appendix Q. A specific attenuation example is presented in Appendix R for Belden 8281 cable, see Fig R.7. However, if losses are so great that the low- loss regime does not apply, the situation becomes much more complicated, and we address that case in a cursory manner below. Basically one cannot consid er the transverse Helmholtz equation as a Laplace equation, so one cannot solve things in the capacitor electrostatics sense, and our rote formulas for K such as those derived in Chapter 6 (like K = 2 ln (a 2/a1) for a coaxial cable) are no longer correct. It turns out that in the high-loss regime K must be determined by solving an unpleasant eigenvalue problem. One might argue that the high-loss regime is of little practical interest since practical transmission lines are always designed to be low-loss transmission lines. Let's go back to our equation before the low-loss assumption that Z s= 0, [ ∇t2 + jω Zs 4πξd K ] φt(x,y) = 0 φt(C1) = K/2 φt(C2) = - K/2 (5.3.10) This is now a Helmholtz equa tion with Helmholtz parameter jω Zs 4πξd K , whereas with Z s = 0 we had the simpler Laplace Equation. Treating Z s as some given value ≠ 0, we could go ahead and find the Green's function for the above equation and it would be a function of K since K appears in the Helmholtz Chapter 5: The Transverse Problem 190 parameter. Call this Helmholtz Green's function g K(x,y|x',y'). We still have φ = 1 4πξd q(z) φt being the full potential from which the electric field is obtained as E n = -∂nφ [ recall that transverse A components are zero so this is consistent with E = - grad φ - ∂tA ]. The solution of the above PDE system then starts off φt(x,y) = – ∫{C1 ds' K1 ∂ngK(x,y|x',y') – ∫{C2 ds' K2 ∂ngK(x,y|x',y') = K 1 F1(x,y,K) + K 2F2(x,y,K) . (5.6.1) From this point on, every function and constant acquires and argument K: G ij(x,K), Hij(K) and then cij(K). We end up then with K = 4πε c11(K) + c22(K) + 2c 12(K) c11(K)c22(K) - [c12(K)]2 . (5.6.2) The new feature is that K appears on both sides of th e last equation. This probably-complicated equation then has to be solved for K, a nd sometimes this is referred to as "an eigenvalue problem" for K. For example, if Z s is very small but non-zero, one would expect the solution for K to be slightly different from the value obtained with Z s = 0 and one could perhaps appro ach the problem using perturbation theory where the Helmholtz parameter is a "smallness parameter". Recall that k 2 = βd2 - jω Zs 4πξd / K (5.3.8) where now K is the "eigenvalue" of our solution above. If Z s is very small but not zero, we end up then with k = βd - Δ where Δ is a small complex number. The longitudinal transmission line behavior of all z-dependent functions like φ , Azt, q, V, W, E, B is then given by (5.1.11), q(z,t) = q(z,t) = q(0) ej(ωt-kz) = q(0) ej(ωt-[βd-Δ]z) = q(0) e j(ωt-[βd-Re(Δ)]z) e–Im(Δ)z The real part of Δ causes a shift in the wavenumber k so the wave no longer propagates with the normal dielectric wavenumber βd. Since v = ω/k, we will find that the wave is "slowed down" due to the drag effect of the non-zero surface impedance of the conductors. The imaginary part of Δ then causes an exponential decay of the wave magnitude due to ohmic losses at the conductor surface. In our Chapter 2 Chapter 5: The Transverse Problem 191 analysis of the round wire we found that in general Z s is itself complex, so computation of Δ is a somewhat complicated problem which we shall not attempt here (but see Appendix Q). The problem of lossy transmission lines is usually approached using E and B fields, rather than potentials φ and A z, and the analysis is then similar to the way wa veguides in general are treated. Due to the skin effect, the E and B fields penetrate a distance ~δ into the conductor surfaces and this results in ohmic losses and a "drag" on the propagating wave. In this approach, one ends up again with an eigenvalue problem to solve, not directly for K but for some other related parameter like k. In the 12-page Section 4.5 of his book, Matick studies a lossy-transmission line in the simplest possible case which is a strip geometry whose gap S is small compared to the width, and whose metal strips are much thicker than the skin depth δ. His parameter γ is related to our parameter k by γ = jk, and his longitudinal direction is x instead of our z. He ends up with a transcendental "e igenvalue equation" (4-66) for γ, but if loss is very small, he can approximately solve for γ with these results [ β d = ω μdεd ] Im(γ) = βd(1+δ/2S) Re( γ) = βd (δ/2S) // Matick (4-75,76,77) p 115 which with γ = jk we translate to Im(k) = - Re( γ) = - β d (δ/ 2 S ) Re(k) = Im( γ) = βd(1+δ/2S) k = βd(1+δ/2S) -j βd (δ/2S) = β d [1 + (δ/2S) + j(δ /2S)] so Δ = β d[(δ/2S) + j(δ /2S)] . Thus, for such a thick strip transmission line, the long itudinal dependence of all functions has this form, q(z,t) = ej(ωt-[βd-Re(Δ)]z) e–Im(Δ)z = ej(ωt-[βd+δ/2S]z) e–(δ/2S)z which shows the exponential loss fact or and an increased wavenumber β+δ/2S which corresponds to a decreased wavelength λ and a decreased wave velocity v = ω/k = ωλ /2π = fλ , the "drag effect". Matick has an erratum in this section which is a bit confusing, so we repair it right here. His equation (4-50) p 110 should read (in his notation) ∇ 2E = (∂2Ex ∂x2 + ∂2Ex ∂z2 ) x^ + (∂2Ez ∂x2 + ∂2Ez ∂z2 ) z^ = (jωμσ - ω2με)(Exx^ + Ezz^) Matick (4-50) Chapter 6: Transmission Lines with Two Cylindrical Conductors 192 Chapter 6: Two Cylindrical Conductors 6.1 A candidate transverse potential φt In the previous chapter (both Section 5.3 (b) and Sec tion 5.4 (b)) we showed that the transverse potential of a 2-conductor balanced transmission line must have this form when viewed from far away, φt(x) ≈ ln(s22/s12) // limiting form as point x = (x,y) moves far from the conductors (5.3.13) s12 = (x-x1)2 + (y-y1)2 = |x - x1|2 s22 = (x-x2)2 + (y-y2)2 = |x - x2|2 ( 6 . 1 . 1 ) where the points x 1 and x2 are the "center of charge" points for the C 1 and C2 conductor cross sections. Suppose now we take as a candidate dimensionless transverse potential φ t exactly the above limiting expression. Our candidate φt is φt(x) = ln(s22/s12) . for all values of r, close and far (6.1.2) where we specify that our center of charge points are x 1 = (d,0) and x2 = (-d,0). Certainly this meets our limiting form boundary condition (5.3.13)! We know also that this potential is a valid solution of the 2D Laplace equation, since ln(s 1) and ln(s 2) are each valid solutions. This fact was shown at the start of Section 5.4 (b). Since -2 πlns1 is the 2D free-space propagator, it follows that -2πlns1 is a solution of ∇2 2D(φ) = 0 away from the point where s 1 = 0, and then so is lns 1. Then by superposition, 2lns 2 - 2lns1 is also a valid solution, and thus so is ln(s 22/s12). Thus, our φt is a valid candidate for a lossless transmission line since for such a transmission line φt satisfies the 2D Laplace equation according to (5.4.3), ∇ t2 φt(x,y) = 0 φt(C1) = K1 φt(C2) = K2 K 1-K2 = K . (5.4.3) The question then becomes: what are the surfaces C i in 2D space on which this candidate φt is a constant? Such surfaces can then serve as possible conductor cross sections for a transmission line. 6.2 Ancient Greece circa 230 BC Apollonius of Perga (262BC-190BC) [ like Joe of Chi cago ] was a pretty smart guy as wiki explains. He did astronomy and therefore he did geometry. Besides giving conic sections their current names and writing eight books about them, he learned about what are now called the Apollonian Circles. These circles form the "level surfaces" for 2D bipolar (o rthogonal) coordinates as shown in this picture, Chapter 6: Transmission Lines with Two Cylindrical Conductors 193 Fig 6.1 http://en.wikipedia.org/ wiki/Apollonian_circles When this picture is rotated around its vertical axis, the blue level circles become toroids and one then arrives at 3D toroidal coordinates, but that is anothe r story. Our interest is in the 2D blue circles. It turns out, as the reader may su spect, that the blue circles have the following simple property, s 2/s1 = c o n s t a n t , which we shall prove in a mome nt. Calling this constant e -B we get s 2/s1 = e-B => ln(s2/s1) = - B . (6.2.1) Thus, since φ t(x) = ln(s22/s12) = 2 ln(s 2/s1), the blue circles are candida te equipotential surfaces for our potential φt(x). To show that s 2/s1 = e-B describes a circle, consider: |x-x 2| / |x-x1| = e-B |x-x 2|2 = e-2B |x-x1|2 (x-x 2)2 + (y-y2)2 = e-2B [(x-x1)2 + (y-y1)2 ] . This equation has the following form Chapter 6: Transmission Lines with Two Cylindrical Conductors 194 A(x2 + y2) + Bx + Cy + D = 0 A = (1-e-2B) or x 2 + y2 + αx + βy + γ = 0 . One can then "complete the squares" to obtain th e equation of a circle of radius r centered at (x c,yc) , (x - x c)2 + (y - yc)2 = r2 where -2x c = α -2yc = β xc2 + yc2 - r2 = γ . (6.2.2) For our particular locations of x 1 and x2 shown in Fig 6.1, we have x 1 = -d x 2 = d y 1 = y2 = 0 s12 = (x+d)2 + y2 s 22 = (x-d)2 + y2 ( 6 . 2 . 3 ) so s 2/s1 = e-B => s1/s2 = eB => e2Bs22 = s12 => (eB/2) s22 = (e-B/2) s12 => (e B/2) [x2 - 2dx + d2 + y2] = (e-B/2) [x2 + 2dx + d2 + y2] shB (x 2+d2+y2) + chB(-2dx) = 0 // shB = (eB-e-B)/2, chB = (eB+e-B)/2 (x2+d2+y2) + cothB (-2dx) = 0 x2 - 2d x cothB + y2 = -d2 x 2 - 2d x cothB + d2coth2B + y2 = -d2+ d2coth2B // complete the square (x - dcothB) 2 + y2 = d2csch2B . ( 6 . 2 . 4 ) We conclude that our blue equipotential circles have this simple form (x - x c)2 + y2 = r2 xc = d cothB r = |d cschB| . (6.2.5) Using d = 5, here is a plot of these circles for 10 different B values: Chapter 6: Transmission Lines with Two Cylindrical Conductors 195 Fig 6.2 Since xc = d cothB, the right side circles have B > 0 wh ile the left side have B < 0. The value B = 0 corresponds to the vertical y axis, while B = ± ∞ correspond to the two focal points at d = ± 5. 6.3 Back to the Future: Calculation of K We sele ct C2 to be a circle on the right side, so that B 2 > 0. For C1 we select a second circle from either the left or the right, so B 1 can have either sign. If we select C 1 from the left side, we have a two-wire transmission line(dielectric = gray), Fig 6.3 If we select C 2 from the right, we have an off-center coaxial transmission line. Chapter 6: Transmission Lines with Two Cylindrical Conductors 196 Fig 6.4 Fig 6.3 shows a transmission line cross section wher e the two conductors are round wires with unequal radii a1 and a2. Treated as a 2D capacitor, one's intuition at least suggests that the two focal points might be the conductor "centers of charge". The gray dielectric is of course outside the two conductors and it is possible to select a point in the dielectric that is "far away" from both conductors, so our limiting form discussion applies and the points x1 and x2 should be the centers of charge. Figure 6.4 shows an off-center coaxial transmission lin e for which the dielectric is the region between the two black circles. In this case, one cannot take a point in the dielectric that is "far away" from both conductors, so the limiting form discussion does not appl y. Here it appears that both conductors have the same center of charge located at x2. We shall now determine K and therefore the 2D capacitance C = 4 πε/K for the above cases. Let σ1 = sign(B 1). We then have from (6.1.2) and (6.2.1), φ t(x) = ln(s22/s12) = 2 ln(s 2/s1) φt(C1) = 2 ln(s 2/s1)|C1 = -2B1 φt(C2) = 2 ln(s 2/s1)|C2 = -2B2 . (6.3.1) Recall from (5.1.3) that φt(C1) - φt(C2) = K. Therefore, K = 2(B 2-B1) = 2 (|B 2| -σ1|B1| ) . ( 6 . 3 . 2 ) Once we know K, we know C, G and L e for the transmission line from box (4.11.34). We must now do some slightly painful algebra. First, we know from (6.2.5) that a 1 = d |cschB 1| => (d/a 1) = sh(|B 1|) => |B 1| = sh-1(d/a1) a2 = d |cschB 2| => (d/a2) = sh(|B 2|) => |B 2| = sh-1(d/a2) . (6.3.3) The separation of the centers of the two round wires is b, where, again using (6.2.5), b = |xc2 - xc1| = |d cothB 2 - dcothB 1| = d |cothB 2 - cothB1| . (6.3.4) Chapter 6: Transmission Lines with Two Cylindrical Conductors 197 From (6.3.2) we write ch(K/2) = ch [|B 2| -σ1|B1|] = ch|B 2| ch|B1| - σ1 sh|B2| sh|B1| = 1+sh2B2 1+sh2B1 - σ1 sh|B2| sh|B1| = 1+(d/a2)2 1+(d/a1)2 - σ1 (d/a2) (d/a1) . (6.3.5) Meanwhile, b = d |cothB 2 - cothB1| = |d [ chB 2/shB2 - chB1/shB1] | = |d [ ch|B 2|/sh|B2| - σ1ch|B1|/sh|B1|] | = | d [ c h | B 2| sh|B1| - σ1 ch|B1| sh|B2| ] / sh|B 1| sh|B2| | = | d [ 1+(d/a2)2 (d/a1) - σ1 1+(d/a1)2 (d/a2) ] / (d/a 2) (d/a1) | = | [ 1+(d/a2)2 (1/a1) - σ1 1+(d/a1)2 (1/a2) ] / (1/a 2) (1/a1) | = | [ a 2 1+(d/a2)2 - σ1 a1 1+(d/a1)2 ] | . (6.3.6) Square this to get b 2 = a22[1+(d/a2)2] + a12[1+(d/a1)2] - 2σ1a1a21+ (d/a2)2 1+ (d/a1)2 so 2 σ 1a1a21+ (d/a2)2 1+ (d/a1)2 = a22[1+(d/a2)2] + a12[1+(d/a1)2] - b2 = a22 + d2 + a12 + d2 - b2 = a12 + a22 + 2d2 - b2 . The purpose of doing this is to obtain the following expression for the radical product, 1+ (d/a2)2 1+ (d/a1)2 = (a12 + a22 + 2d2 - b2) / (2 σ1a1a2 ) . (6.3.7) We now install this into our expressi on (6.3.5) above for ch(K/2) to get ch(K/2) = 1+(d/a2)2 1+(d/a1)2 - σ1 (d/a2) (d/a1) = ( a 12 + a22 + 2d2 - b2) / (2 σ1a1a2 ) - 2d2/ (2σ1a2a1) = ( a 12 + a22 - b2) / (2σ1a2a1) = σ1 (1/2) (a12 + a22 - b2)/(a1a2) Chapter 6: Transmission Lines with Two Cylindrical Conductors 198 = σ1 (1/2) [ (a 1/a2) + (a2/a1) - (b2/a1a2) ] (6.3.8) and the focal distance d has vanished from our expression. Therefore K = 2 ch -1 { σ1 (1/2) [ (a 1/a2) + (a2/a1) - (b2/a1a2) ] } (6.3.9) Notice that the result is symmetric under a 1 ↔ a2 . We now distinguish our two cases of interest. For th e unequal twin-lead type transmission line of Fig 6.3 we know that B 1 < 0 since the C 1 circle is on the left, so σ1 = sign(B 1) = - 1 and then K = 2 ch-1 { (1/2) [ (b2/a1a2) - (a1/a2) - (a2/a1)] } // Fig 6.3 (6.3.10) which is an amazingly simple result. Recall from (4.4.16) that Z0 = (K /εrel ) 30Ω (4.4.16) so then Z0 = ch-1 { (1/2) [ (b2/a1a2) - (a1/a2) - (a2/a1)] } (1/ εrel ) 60 Ω . (6.3.11) If the wires have diameters d 1 = 2a1 and d2 = 2a2 this becomes Z0 = ch-1 { (1/2) [ (4b2/d1d2) - (d1/d2) - (d2/d1)] } (1/ εrel ) 60 Ω . (6.3.12) For verification, we quote again from Reference RDE page 29-23, where our b is called D. On the other hand, if we are interested in an off-center coaxial transmission line as in Fig 6.4, we select C 1 from the right side of Fig 6.2 and then σ1 = sign(B 1) = +1 and we find Chapter 6: Transmission Lines with Two Cylindrical Conductors 199 K = 2 ch-1 { (1/2) [ (a 1/a2) + (a2/a1) - (b2/a1a2) ] } // Fig 6.4 (6.3.13) Z0 = ch-1 { (1/2) [ (a 1/a2) + (a2/a1) - (b2/a1a2) } (1/εrel ) 60 Ω (6.3.14) Z 0 = ch-1 { (1/2) [ (d 1/d2) + (d2/d1) - (4b2/d1d2) } (1/εrel ) 60 Ω (6.3.15) For verification, we quote again from Reference RDE page 29-24, where we may take d = our d 1 and D = our d 2 and c = our b = the center-line separation. There is one more case of interest that falls out from this analysis. If we take B 1 = 0 we have Fig 6.5 which is a transmission line consisting of a round wire a bove an infinite flat plane. This is a tricky limit of (6.3.10) where both a 1→∞ and b→∞ , so we ignore (6.3.10) and work from scratch. Since B 1 = 0 we find from (6.3.2) that K = 2B 2 . ( 6 . 3 . 1 6 ) We know from (6.2.5) that Chapter 6: Transmission Lines with Two Cylindrical Conductors 200 x2c = d coth B 2 = d chB 2/shB2 a2 = d/sh(B 2) . ( 6 . 3 . 1 7 ) Therefore x 2c/a2 = chB2 => B 2 = ch-1(x2c/a2) => K = 2 ch-1(x2c/a2) . (6.3.18) Here x2c is the distance from the wire center line to the ground plane. If we call this h and the wire radius a, we then have the following extremely simple and exact result, K = 2 ch-1(h/a) // wire radius a with center h over ground plane, exact Z0 = (K /εrel ) 30Ω = ch-1(h/a) (1/ εrel ) 60 Ω ( 6 . 3 . 1 9 ) where we must have h > a to keep the wire from touching the ground plane. Using the identity ch -1x = ln(x + x2-1 ) for x ≥ 1 we can write the above as K = 2 ln [ (h/a) + (h/a)2-1 ] // wire radius a with center h over ground plane, exact Z0 = ln [ (h/a) + (h/a)2-1 ] (1/εrel ) 60 Ω ( 6 . 3 . 2 0 ) For h >> a this becomes ("thin wire") K = 2 ln(2h/a) // wire radius a center h over ground plane, h>> a Z 0 = ln(2h/a) (1/ εrel ) 60 Ω . ( 6 . 3 . 2 1 ) For verification, we found the following web offering (where log means ln ), http://members3.jcom.home.ne.jp/zakii/tline_e/14_microstripline_z0.htm which results are derived using an image method to handle the ground plane. Chapter 6: Transmission Lines with Two Cylindrical Conductors 201 For some odd reason, our usual RDE source on this subj ect only gives the result for h >> a . Taking d to be the wire diameter, Z0 = ln(4h/d) (1/ εrel ) 60 Ω = ln(10) log (4h/d) (1/ εrel ) 60 Ω ≈ log (4h/d) (1/ εrel ) 138.2 Ω ( 6 . 3 . 2 2 ) which then compare to RDE p 29-22 , Reader Exercise: Given φ(x) = ln(s22/s12), compute E = - ∇φ , compute E n = E • n^ as the normal electric field at the surface of C 2, compute n = εdEn as the charge density on C 2, then using that n, find the "center of charge" <x> = [ ∫C2 ds x n( x) / ∫C2 ds n( x) ] and see if <x> = d. Decide whether or not it is worth while learning how to work in bipolar coordinates to carry out this exercise. [ The solution to this exercise appears in the author's Bipolar Coordinates document, see References. See also Section 6.5 below. ] Chapter 6: Transmission Lines with Two Cylindrical Conductors 202 6.4 Summary of Line Parameter Results Summary for Transmission Line with Two Round Conductors (6.3.23) Identities : ch-1x = ln(x + x2-1 ) , x ≥ 1 ch-1x ≈ ln(2x), x >> 1 c h-1[ 1 2 (b a +a b ) ] = ln b a b > a > 0 (4.6.6) Line Properties : C = 4πεd/K, G = 4 πσd/K, Le = μd 4π K εd,σd,μd for dielectric (4.11.34) _____________________________________________________________________________________ d i e l e c t r i c i s g r a y K = 2 ch-1 { (1/2) [ (b2/a1a2) - (a1/a2) - (a2/a1)] } a i = radii b = center separation Special case a 1 = a2 = a: K = 2 ch-1 [ (b2/2a2) - 1] (twin-lead) Special case b >> a 1,a2: K = 4 ln(b/ a1a2 ) Special case b >> a 1=a2=a: K = 4 ln(b/a) _____________________________________________________________________________________ K = 2 ch-1 { (1/2) [ (a 1/a2) + (a2/a1) - (b2/a1a2) ] } a i = radii b = center separation Special case b = 0 and a 2> a1: K = 2 ln(a 2/a1) (centered coaxial) _________________________________________________________________________________ K = 2 ch-1(h/a) = 2 ln [ (h/a) + (h/a)2-1 ] a = radius h = height of center over plane Special case h >> a: K = 2 ln(2h/a) (thin wire) Chapter 6: Transmission Lines with Two Cylindrical Conductors 203 6.5. The Proximity Effect for a Tran smission Line made of Two Round Wires This effect is discussed qualitatively in Appendix P in terms of eddy currents, and we quote the following Figure P.13, F i g 6 . 6 The effect is that for ω >0 the current density J z is not uniform in the conductor cross sections but is larger on the side of each conductor which faces the other conductor. In this section we shall compute J z over the wire cross section and pe rimeter to get a quantitative result. (a) The surface charge density and its moments On either of the conductors shown above there is some surface charge density n( θ) which has moments called N m and ηm in Appendix D. Using the electro-quasi-static model for a transmission line, one can analyze the transmission line as if it were an electro statics capacitor problem: the two cylinders form a capacitor (per unit length). If one assumes a potential V between the conductors, one can solve the Laplace equation to get the potential φ in the dielectric between the conductors, which φ will be constant on the surface of either conductor. From this one may co mpute the electric field in the dielectric, and from the electric field just above the conductor surfaces one can compute n( θ). This calculation is carried out in our (downloadable) document Bipolar Coordinates and the Two-Cylinder Capacitor from which we quote results below. Each cylinder of the transmission line is characterized by a certain value of B as shown in Fig 6.2. In Bipolar B is called ξ which is one of the bipolar coordinates ( ξ,u). The angle θ is measured as indicated in this figure taken from Bipolar , which happens to show the two cylinders having the same radius: Bipolar (7.1) Fig 6.7 Chapter 6: Transmission Lines with Two Cylindrical Conductors 204 Notice that the two bipolar "focal points" are at x = ±d , while the radii of the left and right cylinders are a1 and a2. In Bipolar these parameters d, a 1, a2 are called a, R 1, R2. Comment : It is shown in Bipolar Section 10 (d) that the "center of charge" for the surface charge distribution n( θ) is in fact the focal point for each conductor. Here is the more general picture where the cylinders have different radii. The right cylinder has bipolar coordinate ξ 2 > 0 and the left has ξ1 < 0 Bipolar (10.2) Fig 6.8 The angular surface charge densities on the conductors are found to be (Cou/m), n 1(ξ1,θ) = q 2π |shξ1| chξ1+cosθ n2(ξ2,θ) = - q 2π |shξ2| chξ2+cosθ Bipolar (10.28) (6.5.1) where q is q = 2πε d V ξ2-ξ1 Bipolar (10.15) (6.5.2) and εd is for the dielectric between the conductors. Here q is the charge per unit length in z on the left conductor so has dimensions Cou/ m. The surface charge density n1 is normalized so ∫n1(θ)dθ = q so the dimensions of n1 are Cou/m. The true charge density is n 1(θ) = n1(θ)/a Cou/m2. The capacitance per unit length is then Chapter 6: Transmission Lines with Two Cylindrical Conductors 205 C = q/V = 2 πεd 1 ξ2-ξ1 . dim( εd) = farad/m Bipolar (10.16) (6.5.3) This is in agreement with (6.3.2) which says K = 2(B 2-B1) = 2(ξ2-ξ1) and (4.11.34) that C = 4 πεd/K . Notice that for fixed q the charge distribution on each conductor is independent of the ξ value of the other conductor. Thus, if the battery in Fig 6.8 is disconnected, n1(ξ1,θ) does not change if ξ 2 is varied. Using (D.1.5b) the moments of the surface charge distribution n 1(ξ1,θ) are computed in Bipolar Appendix A and are found to be, ηm ≡ Nm/N0 = (-1)m e-|mξ1| . Bipolar (A.12) (6.5.4) Using (D.1.5a) one then finds, n1(ξ1,θ) = (q/2π)[ 1 + 2 ∑ m = 1∞ (-1)m e-m|ξ1| cos(mθ) ] . Bipolar (A.13) (6.5.5) and in Bipolar Appendix A it is verified that this series sums to the expression in (6.5.1). For small ξ1 there are many significant partial waves in the sum. At θ = 0 the partial waves tend to cancel due to the alternating signs of the terms due to (-1)m, whereas at θ = π the terms reinforce. As expected, the charge density peaks on the side of the conducto r facing the other conductor. Here are plots of the charge distribution n1(ξ1,θ) (6.5.1) for various values of ξ1 and for fixed q ( q/2π = 1) : Bipolar (10.40) Fig 6.9 Chapter 6: Transmission Lines with Two Cylindrical Conductors 206 Here are some equations of interest ( also stated in Bipolar (11.3) ), a1 = - d/shξ1 // radius of left circle (6.3.3) a2 = d/shξ2 // radius of right circle (6.3.3) b = d (coth ξ2 - cothξ1) // distance between center lines (6.3.4) (6.5.6) and the inverse equations, d = (1/2b) b2 - (a2+a1)2 b2 - (a2-a1)2 ξ1 = - sh-1 (d/a1) ξ2 = sh-1 (d/a2) . Bipolar (11.10) (6.5.7) The first equation of the second set determines the bi polar focal distance d from the two cylinder radii a 1 and a2 and the distance b between their center lines. For a 1 = a2 = a this says d = (1/2) b2-4a2 . (b) The Proximity Effect Appendix D computes the E fields inside a round wire of radius a in terms of the surface charge moments ηm under the assumption that a wave ej(ωt-kz) is traveling down the wire. We first remind the reader of the parameters involved. From (D.2.2), β' 2 ≡ β2 - k2 (D.2.2) where β = ej3π/4 (2 /δ) = (j-1) / δ = ej3π/4 ωμσ (2.2.30) k = -j zy = -j (R+jωL)(G+jωC) . (5.3.5) (6.5.8) Here β is the wavenumber in the conductor medium shown in (1.5.1c), while k is a low-loss effective wavenumber for the transmission line wave having the form ej(ωt-kz). Although k is a free parameter in Appendix D, it is forced equal to -j zy in Chapter 5 where the Helmholtz equation is separated into longitudinal and transverse parts. This identification k = -j zy is established only for high frequencies ( = low-loss), but can be assumed approximately true at lower frequencies. This subject is discussed in detail in Section D.11 (a), and the high and low ω limits of k are obtained in Appendix Q. At high ω one sees that k ≈ -j(jωL)(jω C) = -jω-LC = +ωLC ≈ ω LeC = ω/vd ≡ βd0 (4.11.28) β'2 ≡ β2 - k2 ≈ -jωμσ - βd02 ≈ -jωμσ = β2 where we identify 1/ LC with v d, the speed of light in the dielectric, as shown in (4.11.28). The fact that β2 = -jωμσ is shown below (D.2.2) to be valid for f << 1018 Hz, so for any reasonable large ω we do have β' = β = ej3π/4 (2 /δ) = (j-1) / δ = ej3π/4 ωμσ . (6.5.9) Chapter 6: Transmission Lines with Two Cylindrical Conductors 207 Now, since n(θ ) is real and an even function of θ for our two-cylinder transmission line, η-m = ηm and from (D.10.4a) the longitudinal field E z(r,θ) is shown to be Ez(r,θ) = (1/4) I R dc (aβ') [ f0(r) + 2 Σ m=1∞ fm(r) ηm cos(mθ) ] . (D.10.4a) Using (6.5.4) for the ηm we then get Ez(r,θ) = I Rdc (βa/4) [ f0(r) + 2 Σ m=1∞ (-1)m e-m|ξ1| fm(r) cos(mθ) ] (6.5.10) and then, since R dc = (1/π a2σ), Jz(r,θ) = σ Ez(r,θ) = (I/πa2)(βa/4) [ f0(r) + 2 Σ m=1∞ (-1)m e-m|ξ1| fm(r) cos(mθ) ] where f m(r) ≡ [ Jm(x) Jm+1(xa) - Jm(x) Jm-1(xa) ] x = β r xa = βa β = ej3π/4 (2 /δ) . (6.5.11) This is the longitudinal current density in the left round conductor (the one with ξ1 < 0) and just the fact that it is not constant in θ shows that we have a proximity effect as illustrated in Fig 6.6 above. We refer to this current density J z as being "asymmetric" as opposed to "uniform". (c) Plots of the Proximity and Skin Effects First, it is helpful to have a plot showing the conduct ors for various values of ξ so one can get a feel for how "fat" the cylinders are relative to their separation distance (same as Fig 6.2), Bipolar (2.5) Fig 6.10 From (6.5.11), using J dc= (I/πa2), Jz(r,θ) = Jdc (βa/4) [ f0(r) + 2 Σm=1∞ (-1)m e-m|ξ1| fm(r) cos(mθ) ] . (6.5.12) Chapter 6: Transmission Lines with Two Cylindrical Conductors 208 In Maple code we first enter all the expressions of interest: J z = (6.5.12) β = (6.5.9) f m = (6.5.11) ηm = (6.5.4) x = βr xa = βa Next, specific parameters are entered (xi1 = ξ 1 = -1 and δ/a = 1/10), The first plot is of |J z(r,θ)| where the axes are r and θ : Chapter 6: Transmission Lines with Two Cylindrical Conductors 209 Fig 6.11 This shows the general peaking of |J z(r,θ)| at θ = π (see Fig 6.7) , but the plot we really want to see is |Jz(r,θ)| displayed over the cross section of the round wire: |J z| Distribution in left round wire for a = 10, δ = 1 and ξ1 = - 1.00 Fig 6.12 Observations: • Skin effect for |J z| appropriate to δ/a = 1/10 is seen in the cross section. The "bottom" of the plot is flat at value 0 and indicates no current in the central conductor region -- all current is in the sheath of thickness ≈ 1 just inside the wire radius a = 10. Chapter 6: Transmission Lines with Two Cylindrical Conductors 210 • The distribution is strongly peaked on the side of the conductor facing the other conductor; this is the proximity effect (currents in opposite directions). • Using the formula given in (P.10.7), R = Rdc E(Jz2) [E(Jz)]2 , (P.10.7) one can compute the effect wire resistance R > Rdc using (6.5.11) for J z. The high and low ω limits of f m and Jz = σEz appearing in Appendix D.10 and D.11 simplify this task if those limits are of interest. We now present a few such plots for different values of δ and ξ1. First, for δ/a = 1/10 and ξ1 = -3 : | J z| Distribution in left round wire for a = 10, δ = 1 and ξ 1 = - 3.00 Fig 6.13 Here, for two equal-radius round wires of ξ1= -3 and ξ2= 3, the wires are "far apart" (see Fig 6.10 above). The proximity effect is still present as shown in the right most picture (larger at x = 10 than at x = -10), but the effect is small. On the other hand, the skin effect is still strongly in evidence, and again the entire current is in a sheath just inside r = a = 10 of thickness about δ = 1 unit. Next is an example with δ/a = 1/2 and ξ 1 = -1 : | J z| Distribution in left round wire for a = 10, δ = 5 and ξ 1 = - 1.00 Fig 6.14 Chapter 6: Transmission Lines with Two Cylindrical Conductors 211 In the central drawing we are looking into the bowl of the distribution from above. Since δ/a = 1/2 now, the skin effect is much less pronounced: J z is no longer 0 in the central region as shown on the right. There is only a shallow "lip" around the bowl edge suggesting some skin effect. On the other hand, the proximity effect is still strong since for ξ1 = -1 and ξ2 = 1 the wires are fairly close together as shown in Fig 6.10. For such wires, using (6.5.6), a/b = a1/b = |cschξ1| cothξ2-cothξ1 = |csch(-1)| coth(1)-coth(-1) = csch(1) 2 coth(1) = 1 2cosh(1) = 0.324 ==> b/a = 3.08 and b/(2a) = 1.54 so the ratio of wire center separation to wire diamet er is about 1.5 (which agrees with ruler measurements made on Fig 6.10 above). The two conductors touch when this ratio drops to 1.0. In the next plot, we have a = 10 and δ = 1, but now we plot the value of |J z| at r = a going around the surface of the conductor for a set of different ξ1 values: F i g 6 . 1 5 These plots of |J z| bear a strong resemblance to the surface charge density plots shown above in Fig 6.9 for the same set of ξ1 values. It is shown in (D.10.15) that in the extreme skin effect regime we have Jz(r,θ) = - (jω) (β/βd0) a r e(1+j)(r-a)/ δ n(θ) large ω (D.10.15) (6.5.13) so we are not surprised to see that J z(r,θ) tracks n( θ) in this manner. In the following section, we derive the above tracking relationship directly from div E = 0 . Chapter 6: Transmission Lines with Two Cylindrical Conductors 212 (d) The relationship between J z(a,θ) and n(θ) obtained from div E = 0 Here, assuming the skin effect regime and making a fe w assumptions, we obtain (6.5.13) directly from the div E = 0 equation and the charge pumping boundary condition, just to provide some intuition about the linkage between J z and n(θ). The charge pumping boundary condition of Appendix D says (r = a means r = a- ε) E r(r=a,θ) = (jω/σ) n(θ) (D.2.24) (6.5.14) so the pattern of n( θ) is directly mapped to E r(r=a,θ) at the surface. But we are interested in E z since our current density of interest is J z = σEz. The condition div E = 0 in cylindrical coordinates reads, ∂ r (r Er(r,θ,z)) + ∂ θEθ(r,θ,z) + r ∂zEz(r,θ,z) = 0 . For r = a we know from (3.7.0) that E θ(a,θ) = 0 and ∂ θEθ(r,θ) = 0, so for r just below the surface we expect ∂r (r Er(r,θ,z)) + r ∂ zEz(r,θ,z) ≈ 0 // near r = a ∂r (r Er(r,θ)) - jβd0 r Ez(r,θ) ≈ 0 // using ∂ z → -jβd0, see (D.1.16), then cancel ejβdz factors Ez(r,θ) ≈ (1/jβd0) (1/r) ∂r (r Er(r,θ)) . // near r = a (6.5.15) For the symmetric-environment round wire, we know from (2.2.29) that E z(r) = Ez(a) J0(βr) J0(βa) . (2.2.29) Taking the large argument limits of the two Bessel functions using (2.3.3) and (2.3.6), we find that in the skin effect regime, E z(r,θ) ≈ Ez(a,θ) a r e(r-a)/δ ej(r-a)/ δ ( 6 . 5 . 1 6 ) which we note has the same general form as the simple result (2.1.8) with x = a-r which is e-x/δ e-jx/δ and is also consistent with (2.3.7) for magnitude. If we blindly assume this same equation applies to Er(r,θ) and Er(a,θ), then Er(r,θ) ≈ Er(a,θ) a r e(r-a)/δ ej(r-a)/ δ . (6.5.17) Then (6.5.15) says, Chapter 6: Transmission Lines with Two Cylindrical Conductors 213 Ez(a,θ) = (1/jβd0) Er(a,θ)[ (1/r) ∂r (r a r e(r-a)/δ ej(r-a)/ δ) ] |r=a = (1/ja βd0) Er(a,θ) [1/2 + (1+j)a/δ ] ( 6 . 5 . 1 8 ) where the derivative is done by Maple, If in (6.5.18) we use the boundary condition (6.5.14) that E r(a,θ) = (jω/σ) n(θ), the result is E z(a,θ) = (1/ja βd0) {(jω /σ) n(θ)} [1/2 + (1+j)a/ δ ] = (1/ja β d0) {(jω /σ) n(θ)} (1+j)a/ δ ] // ignore 1/2 relative to a/ δ = (1/j βd0) {(ω /σ) n(θ)} (j-1)/ δ ] = (1/j βd0) {(ω /σ) n(θ)} β ] // (6.5.8) = (-j ω/σ) (β/β d0) n(θ) . (6.5.19) Putting this into (6.5.16) then gives E z(r,θ) ≈ (-jω/σ) (β/βd0) n(θ) a r e(r-a)/δ ej(r-a)/ δ (6.5.20) which agrees with our earlier result (6.5.13) quoted from Appendix D. Although we just guessed at the form (6.5.17), that form is verified in box (D.10.13). The bottom li ne here is that E z (and thus J z) "tracks" n(θ) for its θ dependence. Chapter 6: Transmission Lines with Two Cylindrical Conductors 214 (e) The Proximity Effect At Low Frequencies As discus sed in Section D.11(a), the Appendix D ansatz that the z dependence of the transmission line fields has the simple z dependence e-jkz is incorrect at low frequencies. This is so because the physics- derived transmission line equations (4.11.15) which imply this z depe ndence are themselves inaccurate at low frequencies. Thus, although we might expect our transmission line theory to be approximately accurate at low frequencies, we should be prepared for incorrect predictions. One such anomaly is noted in (D.11.11) where the theory blindly extended down to very low frequencies (near and at DC) says, E z(r,θ) Ez(r,0) = 1 + Σm=1∞ (r/a)m (m+1) ηm cos(mθ) (D.11.11) and for the two-cylinder transmission line (6.5.4) gives E z(r,θ) Ez(r,0) = 1 + Σm=1∞ (r/a)m (m+1) (-1)m e-|mξ1| cos(mθ) . (6.5.21) In the DC limit, and very close to it, we expect the longitudinal current J z to be completely uniform across the conductor cross section (non-conducting dielectric), yet the above expression says it is non-uniform since it is a function of θ. For closely spaced conductors (small ξ 1), the predicted non-uniformity is quite dramatic and is sim ilar to plots shown earlier. We know that for the problem of two parallel cylindrical conductors (or any uniform parallel conductors) which carry I and -I (perhaps they are shorted together at one end) , J z is uniform at DC. We know this because at ω = 0 there are no eddy currents induced by on e wire into the other. The DC B field of wire #2 has no influence on the current density distribution J z(r,θ) in wire #1. It does induce a tiny Hall charge onto the surface of wire #1 and a corresponding tr ansverse Hall E field, since the B field of wire #2 temporarily deflects electrons in wire #1 (see Appendix N for various Hall examples). This tiny deflection effect is mentioned in the text below Fi g P.12 in Appendix P, but there is no effect on the J z distribution. From a current density standpoint, J z in wire #1 doesn't even know that wire #2 is present, so wire #2 could just as well be removed. The isolated wire #1 then if round (Chapter 2) would certainly have a current distribution at DC that was independent of θ. So we accept that our theory makes this incorrect prediction as ω→ 0 and we chalk it up to the expected inaccuracy of the theory at low ω . This anomaly is somewhat soft ened when we remember that our theory only applies to infinite transmission lin es, or transmission lines terminated in the correct Z 0. As ω→0, that correct Z 0 → ∞ (non-conducting dielectric) and the current in the wire I → 0. In the case of a conducting dielectric, we might expect a non-uniformity in J z. The theory prediction is (D.11.14) in this case and the accuracy of this prediction is left as an unresolved Reader Exercise in the following text. In a proper treatment of the fixed-load finite-length two-cylinder transmission line problem, as ω→0 one would see the eddy currents gradually decr ease, one would arrive at a DC current I ≠ 0, and one would have J z → uniform. A solution to this problem could be based on the eddy current methods outlined qualitatively in Appendix P and is no doubt available somewher e in the literature. This solution would then show the proximity and skin effects gradually vanishing as ω→ 0, leaving a uniform J z. Chapter 6: Transmission Lines with Two Cylindrical Conductors 215 (f) Active perimeter p and Z s for a two-cylinder transmission line This active perimeter p was roughly illustrated for same-radius cylinders in Fig 2.16, Fat twinlead Fig 2.16 First, recall these high-frequency results for such a transmission line, E z(a,θ) = (-jω/σ) (β/βd0) n(θ) // E z just below the surface (6.5.19) n(θ) = (-q/2π)(1/a) shξ2 chξ2+cosθ (6.5.1) = (-q/2 π)(1/a)[ 1 + 2 ∑ m = 1∞ (-1)m e-mξ2 cos(mθ) ] (6.5.5) where we use ξ2 > 0 for the right conductor in Fig. 6.7 and the fact that n = n/a. For this right conductor, these expressions take their maximum values when θ = π : Ez(a,θ)max = Ez(a,π) = (-jω/σ) (β/βd0) n(π) n( π) = (-q/2π)(1/a) shξ2 chξ2-1 (6.5.22) Meanwhile, the average values are < E z(a,θ)> = (-jω /σ) (β/βd0) <n(θ )> with <n(θ )> = (-q/2 π)(1/a) . // (D.1.8) with q→ -q (6.5.23) Thus, < E z(a,θ)> Ez(a,θ)max = <n(θ)> n(π) = chξ2-1 shξ2 . ( 6 . 5 . 2 4 ) Our active perimeter distance p is defined in (4.11.10) as [ using Z s(θ) = Ez(a,θ)/ I ] p ≡ <Zs(θ)> Zsmax P = <Ez(a,θ)> Ezmax P = chξ2-1 shξ2 2πa . // P = 2π a (6.5.25) Using shξ 2 = ( d / a ) (6.5.6) b = 2d coth ξ2 = 2d chξ2/shξ2 (6.5.6) Chapter 6: Transmission Lines with Two Cylindrical Conductors 216 d = (1/2) b2-4a2 (6.5.7) we find that chξ2 = (b/2d) sh ξ2 = (b/2d)(d/a) = b/2a . (6.5.26) Therefore p = chξ 2-1 shξ2 2πa = (b/2a)-1 (d/a) 2πa = b-2a 2d = b-2a b2-4a2 2πa = b-2a b+2a 2πa = 1-2a/b 1+2a/b 2πa = 2π ae where a e = a 1-2a/b 1+2a/b . (6.5.27) We next compute the average Z s quantity defined in (4.11.9) : Z s ≡ <Zs(θ)> = < E z(a,θ)>/I = (-jω /σ) (β/βd0) (-q/2π)(1/a)/I . (6.5.28) But using (4.11.19a) that I = (-q) v d along with βd0 = ω/vd and β = (j-1)/δ, this becomes Z s ≡ <Zs(θ)> = -(jω/σ) (j-1)/δ * (vd/ω) (1/a) (-q/2 π) (-1/qvd) = -(j/ σ) (j-1)/δ * (1/a) (1/2 π) = (-j) (j-1) (1/2 πaσδ) = 1 2πaσδ (1 + j). (6.5.29) Note that this is the same high- ω result found in (2.4.16) for the axially symmetric round wire situation. In the closely-spaced transmission line, Z s(θ) is very large near θ = π, but is very small near θ = 0, but the average is the same as for a wide-spaced transmission line where each round wire is effectively in isolation. This average Z s is what we use in (4.11.17) to evalua te the transmission line parameters. This calculation already accounts for the "proximity effect" sin ce that effect is built into the theory as shown in the various plots of section (c) above. One can say that Zsmax = Zs P p = Zs a ae = 1 2πaσδ (1 + j) a ae ( 6 . 5 . 3 0 ) which perhaps is the meaning of King's equation (46) quoted below. On page 30 of TLT King discusses the notion of an effective radius a e but his expression for a e is ae = a 1-(2a/b)2 King (45) and it is not clear how his perimeter 2π ae is defined. King's expression for a e at least agrees with ours in the two important cases b →∞ (ae = a) and b →2a (ae = 0). He is quoting work from other people in this section, and we leave the reader to ponder King's comments directly. His "internal impedance" zi is the same as our "surface impedance" Z s. [We were unable to access King's two references. ] Chapter 6: Transmission Lines with Two Cylindrical Conductors 217 † Chapter 6: Transmission Lines with Two Cylindrical Conductors 218 (g) The Proximity Effect For Currents in the Same Direction Our theory does not model this situation. If the currents in the two conductors are in the same direction, we don't even have a transmission line. However, one could regard two such conductors as the central conductors of a coaxial cable with a distant return sheath : Fig 6.16 We then have a transmission line, and it has an asso ciated "capacitor problem" which one could solve to determine the potential, the E field, and finally the surface charge n( θ) on each central conductor. From that one could compute the moments η m and from that the current distributions J z(r,θ) in the two wires using the methods given above. Even if the two central conductors touch, the problem is well defined and non-singular (unlike our regular twin-lead transmis sion line problem). We know from Fig P.12 that the currents will be largest ne ar the surfaces most distant from the other conductor. Conceptually this can be regarded as just the skin effect applied to the composite central conductor. Multiple central conductor strands could be treated in principle in the same manner. This is the subject of the paper by Smith mentioned in Appendix P.9. Appendix A: Gauge Invariance 219 Appendix A: Gauge Invariance Here we show why it is that, in choosing potentials φ and A, one is allowed to set the divergence of the vector potential A equal to an arbitrary function. Roughly speaking, this freedom of setting div A is called gauge invariance. Our step-by-step approach here is somewhat unconventional and brings in the notion of a Green's function and the particular solution of the Poisson Equation . Some extracurricular topics are brought up which may or may not interest the reader. Each section builds on the previous section. A.0 The Poisson Equation and its Solution Fact 0 : The P oisson Equation - ∇2φ = ρ/ε0 with φ(∞) = 0 has a unique solution as stated below. (A.0.0) For electro statics in an isotropic medium equations (1.1.3) and (1 .1.6) indicate that div E = ρ /ε while (1.1.2) says that curl E = 0. Since curl grad f = 0 for any function f, if one lets E = - grad φ , then curl E = -curl grad φ = 0 ρ/ε = div E = - div grad φ = -∇ 2φ => - ∇2φ = ρ/ε , an equation known as the Poisson Equation . The problem of electrostatics (" potential theory") is then to solve - ∇2φ = ρ/ε for the potential φ, and then E = - grad φ produces the resulting electric field. For a static physical situation (nothing varies w ith time t), the electrostatic potential φ matches the scalar potential φ appearing in (1.3.1). Here ρ(x) refers to the electric charge density. (a) Imagine some static charge distribution ρ(x) that is constrained to a localized region near the origin within infinite space. The distribution ρ(x) includes all charges in this region. Here are some types of charges which would be included in ρ(x): • point charges which are "glued do wn" to certain points in space. • linear continuous charge densities that are glued down along curved filaments in space or which are stable on conducting filaments. • surface charge densities that are either glued to cer tain surfaces, or which are stable because they lie on the surfaces of pieces of conductor (like metal). • 3D continuous charge densities that are glued down in 3D space so they cannot move, or which manage to achieve a stable configuration as free charge (if that is possible!) • surface polarization charge densities not already accounted for by the ε in -∇ 2φ = ρ/ε . By including all these types of charge in ρ , we are able to avoid the complicating issue of "boundary surfaces" in our discussion below, and our only boundary of interest is The Great Sphere which is a sphere of infinite radius surrounding our localized region of interest. From Coulomb's Law (in SI units, and in an isotropic medium of dielectric constant ε) we know that the electric potential φ of a point charge q located at point x' is φ (x) = q/[4πεR] where R = | x-x'| is the distance between charge q at x' and an observation point x. Such a point charge is described by ρ(x) = qδ(x-x'). The general equation which relates φ(x) to ρ (x) is the Poisson Equation, Appendix A: Gauge Invariance 220 - ∇2φ(x) = ρ (x)/ε φ(∞) = 0 . (A.0.1) By including the condition φ(∞) = 0, we are really describing a Poisson "boundary value problem". We add φ(∞) = 0 because we are assuming that ρ(x) is localized as just noted. Both the PDE and the boundary condition are linear. Letting φ = αφ1+ βφ2, and ρ = αρ1+ βρ2, ∇2φ = ∇2(αφ1+ βφ2) = α ∇2φ1 + β∇2φ2 = α ρ1(x)/ε +β ρ2(x)/ε = [α ρ1(x) +β ρ2(x)]/ε = ρ/ε φ(∞) = αφ1(∞) + βφ2(∞) = 0 + 0 = 0. // boundary condition is linear Therefore, we may superpose the potentials of mu ltiple charges to get the potential resulting from a distribution of charges. Thus, we at once obtain this superposed version of Coulomb's Law, φ(x) = 1 4πε ∫d3x' ρ(x') |x-x'| . (A.0.2) Here d3x' ρ(x') = dq(x ') is a differential chunk of charge located at x' contained in tiny volume d3x'. Thus, (A.0.2) must be a solution of (A.0.1 ). If we allow the observation point x to move right on top of some point charge in the distribution ρ, we will get φ = ∞, so we generally avoid such observation points. (b) We would like to explicitly show that (A.0.2) is a solution of (A.0.1) for a general distribution ρ. To this end, we digress to consider th e following equation and its solution, - ∇2g(x,x') = δ(x-x' ) with g( ∞,x') = 0 => g( x,x') = 1 4π 1 |x-x'| . (A.0.3) The equation on the left is Poisson's Equation where ρ consists of a positive point charge of q = ε units sitting at position x'. Recall from above that ρ (x) = qδ(x-x') for a point charge. Coulomb's Law gives the solution shown on the right. Therefore it must be true that - ∇ 2{ 1 |x-x'| } = 4π δ(x-x' ) . ( A . 0 . 4 ) This last equation is derived in Appendix H (see H. 1.4), but we have already shown it is true, given Coulomb's Law. We can now show that (A.0.2) is a solution of (A.0.1) for an arbitrary distribution ρ as follows: - ∇ 2φ(x) = 1 4πε ∫d3x' ρ(x') {- ∇2 1 |x-x'| } = 1 4πε ∫d3x' ρ(x') 4π δ(x-x') = ρ (x)/ε . QED. The assisting function g( x,x') has various names with respect to (A.0.3): the Green's Function or Green function, the fundamental solution, the free-space propagator , or the kernel. It is nothing more than the potential created by a point charge of ε units located at x' and viewed from x . Some authors put a 4 π in Appendix A: Gauge Invariance 221 front of the δ in the left equation of (A.0.3) which causes the 1/(4 π) to be absent in the right equation of (A.0.3). (c) We have found the particular solution of (A.0.1) given by (A.0.2). There are many other PDE solutions which can be obtained by adding to the solution (A.0.2) a solution of - ∇ 2u = 0. This last equation, usually written ∇2u = 0, is called the Laplace Equation, and it is the "homogeneous" form of the Poisson Equation, that is, the right side of the Po isson Equation is set to 0. Solutions u are called homogeneous solutions. One obvious solution is u = 2, so we could then add 2 to (A.0.2) and get a new solution to (A.0.1). Since we have specified that our charge distribution ρ(x) is localized to some region of space, we expect that as x → ∞, we must have φ → 0, as (A.0.1) states. The solution (A.0.2) meets this requirement, but if we add 2, then our boundary condition φ(∞)= 0 is not met, so we must rule out adding a 2. We would also rule out 2x + 3, for example, or 7xy. Recall that ∇2 = ∂x2+ ∂y2+ ∂z2. It turns out that the only solution of ∇2u = 0 which meets the requirement u→0 as x→∞ in all directions is the trivial function u( x) = 0. In 2D one intuitively sees th is because a massless taut thin rubber sheet (drum head) tied down to height u = 0 around a circular perimeter is going to be a flat rubber sheet with u = 0 everywhere. Th e solutions to the 3D equation ∇2u = 0 are called harmonic functions, and it is not hard to show that any harmonic function must take both is max and min values on the boundary, which here is a 3D great sphere. Thus u max = 0 and u min = 0, so the only possibility is that u(x) ≡ 0 everywhere. The implication of the previous paragraph is th at (A.0.2) is the only possible solution of (A.0.1) because the only homogeneous solution one is allowed to add to (A.0.2) is u = 0. One can suppose there are two different solutions of - ∇ 2φ = ρ/ε called φ and φ' both of which go to 0 on the great sphere. Then -∇2(φ-φ') = 0 with ( φ-φ') → 0 on the great sphere. But then ( φ-φ') = 0 so φ ' = φ and there cannot then exist two different physical solutions of (A.0.1). (d) In the following discussions, we shall be less explicit about boundary conditions like φ(∞) = 0, but they are always implied because we shall always be considering only a local distribution of sources. One convenient implication of such boundary conditions is that the "parts" of parts integrations often vanish, since they involve functions evaluated on the Great Sphere (or Great Circle in 2D). To clarify this perhaps obscure comment, here is a statement of two integral theorems where V is an n dimensional volume and S is an n-1 dimensional surface enclosing that volume: ∫V dV ∇φ = ∫S dS φ // "integral of a gradient theorem" (A.0.5) ∫V dV ψ(∇φ) = – ∫V dV (∇ψ)φ + ∫S dS (ψφ) // "parts integration" (A.0.6) "the minus sign" "the parts" The first theorem is just the dive rgence theorem (1.1.30) applied to F (x) = φ(x) a where a is a constant vector. It happens that ∇•F = ∂ iFi = ∂i[φ ai] = (∂iφ) ai = ∇φ • a , so ∫V dV ∇•F = ∫S dS•F => ( ∫V dV ∇φ) • a = ∫S dS•[ φa] = (∫S dS φ) • a Appendix A: Gauge Invariance 222 By setting a = x^, y^ and z^ one concludes that ∫V dV ∇φ = ∫S dS φ. The second theorem is the first applied to the function φ→ψφ and is the generalization of 1D parts integration to n dimensiona l space. The "parts" is ∫S dS (ψφ) and if S is the Great Sphere, then this integral involves ψ and φ evaluated on the Great Sphere, and usually one of these functions is 0 there. In what follows, we shall often be swinging a derivative from one function to the other inside an integral, and we ignore the parts for the reason just stated. A.1 Existence of A such that B = curl A and div A = 0 Fact 1 : If div B = 0, there exists an A s uch that B = curl A and div A = 0. (A.1.0) The "gauge choice" div A = 0 is known as the Coulomb or Transverse Gauge. More on gauges later. Proof : There are several parts to the proof: (a) If A exists such that B = curl A, then it will certainly be true that div B = 0, since div curl A = 0 for any vector field A. The problem is showing that A exists, and moreover, that an A exists with div A = 0. (b) Consider the following differential equation (at this point A is some undefined vector field): - ∇ 2A = curl B (A.1.1a) or, in Cartesian coordinates, - ∇ 2(Ai) = [curl B]i . ( A . 1 . 1 b ) We may regard this as the Poisson equation (A.0.1) where φ → Ai and ρ → ε [curl B] i . We know that a Poisson equation of the form (A.0.1) has a unique physical solution of the form (A.0.2), so the solution of (A.1.1) is given by A(x) = 1 4π ⌡⌠ d3x' curl' B(x') |x - x'| . (A.1.2) As with ρ in the previous section, we think of curl B as being localized in some region near the origin and dropping off at large distances. Perhaps B is generated by some currents in this localized region. We take (A.1.2) to be a candidate expression for the vector field A. If we can show that div A = 0 and that B = curl A , then (A.1.2) is a viable expression for A. Appendix A: Gauge Invariance 223 (c) Take the divergence of both sides of (A.1.2) [ implied sum on i ] div A(x) = ∂iAi(x) = 1 4π ⌡⌠ d3x' [curl' B(x')]i ∂i 1 |x - x'| . (A.1.3) We can replace ∂i by - ∂ 'i acting on 1/| x - x'| . Then we can do parts integration and move ∂ 'i onto [curl' B(x')]i with a parts sign change. In doing so, we assume th at at infinity we pick up no "parts" since curl B is assumed to drop off sufficiently fast. We end up then with: div A(x) = 1 4π ⌡⌠ d3x' div' curl ' B(x') |x - x'| . (A.1.4) But div curl F = 0 for any vector field F , so the integrand and integral vanish. Thus, we conclude that div A = 0 . ( A . 1 . 5 ) (d) Next, take the curl of both sides of (A.1.2). Here is the ith component [ implied sums on j and k, and εijk is the totally antisymmetric permutation tensor used to express curl components ] [curl A(x)]i = εijk∂jAk(x) = + ε ijk 1 4π ⌡⌠ d3x'[curl' B(x')]k ∂j 1 |x - x'| . (A.1.6) As before, replace ∂j by -∂'j acting on (1/| x - x'|). Then do parts to move ∂'j onto [curl' B(x')]k. As before, there is no "parts contribution". The result can th en be put back into full vector notation to give: curl A(x) = + 1 4π ⌡⌠ d3x' curl' curl' B(x') |x - x'| . (A.1.7) Now use the vector identity curl curl B = grad div B - ∇2 B = - ∇2 B , since div B = 0. This gives curl A(x) = - 1 4π ⌡⎮⌠ d3x'∇'2B(x') |x - x'| . ( A . 1 . 8 ) The next step is to move the operator ∇'2 onto the other integrand factor 1/| x - x'| by doing a double parts, and again for each parts operation there is no parts cont ribution from the Great Sphere at infinity. We then use the fact (A.0.4) that ∇2(1/|x - x'|) = - 4π δ(x-x' ) to get curl A(x) = - 1 4π [ -4π Β(x) ] = Β(x) . (A.1.9) Thus, assuming div B = 0, we have formally constructed in (A.1.2) a vector field A such that B = curl A and div A = 0, and this was the claim of Fact 1 stated above. Appendix A: Gauge Invariance 224 A.2 Existence of A' such that B = curl A' and div A' = f . Fact 2 : If div B = 0, there exists A' such that B = curl A' and div A' = f( x), where f( x) is an arbitrary scalar field which "drops off" in some reasonable (sufficient) manner as | x| → ∞. (A.2.0) Proof : From Fact 1, we first find A such that B = curl A and div A = 0. We then define A' ≡ A + grad Λ dim(Λ) = volt-sec (A.2.1) where Λ is some so-far arbitrary function (scalar field). As shown below (1.3.1), dim( A) = volt-sec/m, and therefore dim( Λ) = volt-sec. It follows from (A.2.1) that div A' = div A + ∇ 2 Λ = ∇2Λ . (A.2.2) We would like to have div A' = f, so we must find Λ such that ∇2Λ = f . dim(f) = volt-sec/m2 ( A . 2 . 3 ) But this is once again Poisson's Equation (A.0.1) with φ → Λ and ρ → -εf. Translating (A.0.2) we then find that Λ(x) = – 1 4π ⌡⌠ d3x' f(x') |x - x'| . (A.2.4) Meanwhile, from (A.2.1) we also conclude that, since curl grad g = 0 for any function g, curl A' = curl A + curl grad Λ = curl A = B . (A.2.5) Thus, assuming div B = 0, we have formally constructed a vector field A' such that B = curl A' and div A' = f(x) where f( x) is any function we like that drops off sufficiently fast as | x|→ ∞, and this is the claim of Fact 2. If f(x) drops off away from the origin, this is like the ρ(x) of Fact 0, and we find that Λ → 0 as x → ∞ in any direction. Then since Λ = 0 on the Great Sphere, we know that there are no homogenous solutions to ∇ 2Λ = 0 which could be added to (A.2.4) and so Λ(x) is uniquely determined by our selected function f( x). The function Λ(x) is called a gauge function for reasons given below. A.3 Existence of φ su ch that E = -grad φ Fact 3 : If curl E = 0, then there exists a φ such that E = - grad φ . (A.3.0) Proof : This proof is almost identical to that of Fact 1, but a little simpler. (a) If φ exists such that E = - grad φ, then it will certainly be true that curl E = 0, since curl grad φ = 0 for any function φ. The problem is showing that φ exists. (b) Consider the following differential equation (at this point φ is some undefined scalar field): Appendix A: Gauge Invariance 225 ∇2φ = - div E . (A.3.1) This is yet again Poisson's equation (A.0.1) for φ, this time with ρ→ ε div E, we solve it as in (A.0.2) to get, φ(x) = 1 4π ⌡⌠ d3x' div' E(x' ) |x - x'| . ( A . 3 . 2 ) As usual, we assume that div E drops off in some sufficient ma nner away from the origin going to infinity. Perhaps E is generated by a charge distribution in some region near the origin. (c) Next, take the grad of both sides of (A.3.2). Here is the ith component: ∂iφ(x) = 1 4π ⌡⌠ d3x' div' E (x') ∂j 1 |x - x'| . (A.3.3) As usual, replace ∂ j by -∂'j acting on (1/| x - x'|). Then do parts to move ∂'j onto div' E(x') with a second sign change, and also as usual there is no "parts contribution" from the Great Sphere. The result can then be put back into full vector notation to give: grad φ(x) = + 1 4π ⌡⌠ d3x' grad' div' E(x') |x - x'| . (A.3.4) Now use the vector identity grad div E = curl curl E + ∇2 E = ∇2 E , since curl E= 0. This gives grad φ(x) = 1 4π ⌡⎮⌠ d3x'∇'2E(x') |x - x'| . (A.3.5) As before, move the operator ∇ '2 onto the other term 1/| x - x'| by doing a double parts. We then use the fact (A.0.4) that ∇2(1/|x - x'|) = - 4π δ(x-x' ) to get grad φ(x) = 1 4π [ - 4π Ε(x) ] = −Ε(x) (A.3.6) Thus, assuming curl E = 0, we have constructed a function φ in (A.3.2) such that E = - grad φ, so φ must exist, and this is the claim of Fact 3. A.4 Existence of A' and φ ' such that B = curl A', E = - grad φ ' - ∂tA', and div A' = f. Fact 4 : If div B = 0 and curl E = - ∂B/∂t , then there exist both A' and φ' such that B = curl A' and E = - grad φ ' - ∂A'/∂t , and the quantity div A' may be set to any function f. (A.4.0) Appendix A: Gauge Invariance 226 Proof : We know from Fact 2 that A' exists such that B = curl A' and such that div A' equals any arbitrary function f. If we start with some arbitrary A and φ, the successful A' from (A.2.1) is A' = A + grad Λ where Λ is given by (A.2.4) as an integral over f. What is the corresponding φ' ? Since the E field corresponding to ( A,φ) and ( A',φ') must be the same, we must have - E = -E ' or grad φ + ∂tA = grad φ ' + ∂tA' . Since A' = A + grad Λ, then ∂tA' = ∂tA + grad ∂t Λ, so the above reads grad φ = grad φ ' + grad ∂ t Λ which is satisfied by φ' = φ - ∂ tΛ. Thus, the successful potential pair giving div A' = f is this: A' = A + grad Λ φ' = φ - ∂tΛ / / d i m ( Λ) = v o l t - s e c ( A . 4 . 1 ) where from (A.2.4), Λ(x) = – 1 4π ⌡⌠ d3x'f(x') |x - x'| (A.2.3) The pair of equations (A.4.1) is called a gauge transformation and we have just seen in Facts 2 and 4 that a gauge transformation preserves both E and B. That E' = E was built into (A.4.1), and B' = B since B' = curl A' = curl A + curl grad Λ = B + 0 = B. Each possible choice f defines a function Λ which then gives a transformation. There are an infinite set of f and corresponding Λ functions, so there are an infinite number of gauge transformations which leave the E and B fields invariant. We are free to choose a gauge such that div A' = f for any f we like. A.5 Gauge Invariance In electro magnetism, the situation of Fact 4 arises for B = magnetic field A = vector potential E = electric field φ = scalar potential Using the gauge transformation (A.4.1), one transforms from A,φ to A ',φ' without altering the physical electromagnetic fields E and B. The electromagnetic fields are thus invariant under such a gauge transformation, and one says that the cl assical theory of electromagnetism is gauge invariant . The word "gauge" was first used by Hermann Weyl in the context of general relativity. Gauge invariant there means that a certain "covariant derivative" transf orms as a proper tensor object so that things have the same form in different coordinate systems used to measure things. These diffe rent coordinate systems were referred to as different "gauges" in the sense th at a gauge is a marked-off measuring instrument used Appendix A: Gauge Invariance 227 to measure something (like the marked -off x-axis of a coordinate system). In general relativity the metric tensor g μν, which defines the meaning of distance in the 4 dimensions of spacetime, is a function g μν(x) of the local location in spacetime x. Weyl considered the effect of rescaling the metric tensor according to gμν(x) → λ(x)gμν(x) where λ(x) was an arbitrary "gauge function" ( like our Λ(x) ). Nowadays, gauge invariance is associated with any continuous degree(s) of freedom of a theory which don't affect physical measurements derived from the theory, such as our gauge transformation (A.4.1). See Quigley. A.6 The Lorenz Gauge and QED This section is certainly off the transmission-lines beat en path, but the author thought the reader might find it interesting. It is true that the nature of a transmission line results from photons "jumping back and forth" between the conductors. Unlike elsewhere in this document, everyt hing is not fully explained in the following quick outline. A more deta iled description of the tensor nota tion used below may be found in the author's Tensor Analysis document and elsewhere. In relativistic notation one uses 4-vectors whic h have one time component and three spatial components such as xμ = (ct,x,y,z) which denotes a point in "spacetime". The time component t is multiplied by the speed of light c so that all four components have the same units -- distance L. Often people measure distance in light-seconds instead of meters so in such units c = 1, but we shall display the c to keep track of units. This xμ is a "contravariant" (index up) 4-v ector and the corresponding "covariant" (index down) 4-vector is x μ = (ct,-x,-y,-z). Thus, one has x0 = x0 (= ct) but xi = -xi. We are assuming here the "Bjorken-Drell metric" g μν = diag(1,-1,-1,-1). The gradient operator ∂i "transforms as" the spatial part of the covariant 4-vector ∂μ, and one can write ∂i = -∂i just as xi = -xi for i = 1,2,3. This four-vector gradient operator can be written ∂μ = (∂0, ∂i) and ∂μ = (∂0, ∂i) = (∂0, -∂i) where ∂0 = ∂0 = 1 c ∂t = 1 c ∂ ∂t . The four components of ∂μ all have dimension L-1. The Laplacian is ∇2 = ∂i∂i = ∂ ∂xi ∂ ∂xi (implied sum on i) while the corresponding object … ≡ ∂μ∂μ = 1 c2 ∂t2 - ∇2 is the D'Alembertian which appears in wave equations. Consider then the gauge transformation (A.4 .1) which in relativistic tensor notation is A'i = Ai + ∂iΛ = Ai - ∂iΛ i = 1,2,3 φ' = φ - ∂tΛ = φ - c ∂0Λ . ( A . 6 . 1 ) The components of a classical vector like A, normally written as A i, are in fact the contravariant components Ai in tensor notation. If we now define A0 ≡ 1 c φ we can combine the two gauge transformation equations into a single equation involving three 4-vectors (one of which is ∂μΛ), A' μ = Aμ - ∂μΛ μ = 0,1,2,3 . (A.6.2) Suppose we want ∂ μA'μ = 0 (implicit sum on μ = 0,1,2,3). This would be a relativistic version of the Coulomb gauge choice that ∂iAi = div A = 0. If we could find a potential A'μ with this property, that would be very convenient for the following reason: In general a μbμ (= aμbμ = a • b) is the same in all Appendix A: Gauge Invariance 228 frames of reference related by Lorentz Transformations. If ∂μA'μ = 0 in one frame, it is 0 in all frames, and that makes computational life simple. For example, let S and S" be two frames of reference related by a Lorentz transformation. Then the implication is that ∂μA'μ(xν) = 0 ⇒ ∂"μA'μ(x"ν) = 0 where ∂μ ≡ ∂/∂xμ and ∂"μ ≡ ∂/∂x"μ frame S observer frame S" observer So, is it possible to have ∂μA'μ = 0 ? Writing this out we get ∂0A'0 + ∂iA'i = 0 => 1 c ∂t [1 c φ'] + div A ' = 0 => 1 c2 ∂tφ' + div A' = 0 so div A' = - 1 c2 ∂tφ' . ( A . 6 . 3 ) But we showed in Fact 2 that given any A, we can find an E-B-fields-equivalent A' which has div A' = any f(x ) we want, so we just select f(x) = -(1/c2) ∂φ'/∂t. By selecting this f(x), we are selecting the Lorenz Gauge. In this gauge (now dropping the prime on A), we have ∂μAμ = 0. Thus, the condition defining the Lorenz Gauge is Lorent z covariant under all Lorentz transformations. The reason is that both sides of ∂μAμ = 0 "transform" as the same kind of tensor object, in this case a scalar object. One can interpret ∂μAμ = 0 as ∂ •A = 0 where ∂ is a 4-divergence operator. Thus, in the Lorenz gauge, the 4- divergence of Aμ is always exactly 0 at every point in spacetime. [ Lorenz and Lorentz are two different people, see the Comment below equation (1.3.6).] In 3D if we said that div F = ∂iFi = 0 defined something called a ga uge condition, it would be clear that F was not uniquely determined by that condition since many vector fields have zero divergence. Just so, the Lorenz gauge condition ∂μAμ = 0 does not uniquely determine Aμ , it is just a condition on Aμ. So in fact there are many pairs ( A,φ) which satisfy the Lorenz gauge condition, so the term " the Lorenz gauge" is a little misleading, though we sha ll use it anyway. It is a class of gauges. In relativistic quantum field theory (aka qua ntum electrodynamics, or QED), the potential Aμ is interpreted as the quantum field of a massless vector particle called the photon. The potentials φ and A are thus promoted from being mere "helper functions" to having their own particle interpretation. In the Lagrangian density for the photon-electron system an interaction term - JμAμ appears, L = ... - JμAμ J μ = e0 ψ¯ γμ ψ (A.6.4) where J μ is the electric current, an operator built from the quantum field ψ of the electron. The number e 0 is the so-called bare (unrenormalized) charge of the electron. According to (A.6.2), a gauge transformation on A μ creates a new term - J μ ∂μΛ in the Lagrangian density. In Lagrangian dynamics, the physics of QED is determined by S = ∫d4x L = ∫d3x ∫ dt L which is called the action. If we insert the gauge term -J μ∂μΛ into the action and do parts integration to move ∂μ from Λ to Jμ, we end up with an action change ΔS = ∫d4x (∂μJμ)Λ. But at every point in spacetime, we know that ∂μJμ = 0 (shown in a Appendix A: Gauge Invariance 229 moment) so we find that ΔS = 0 which means the action S is invariant under any gauge transformation. The reason ∂μJμ = ∂μJμ = 0 is because Jμ = (cρ, Ji) where ρ is charge density and Ji is electric current, and then the statement ∂μJμ = 0 says that 1 c ∂t(cρ) + ∂iJi = 0 or div J = -∂ρ/dt. This is the equation of continuity (1.1.8) which says that if there is a current flowing out of a tiny volume of space, the charge density in that volume must be correspondingly decreasi ng. In other words, charge is "conserved". We can reverse our logic to conclude that the reason elec tric charge is conserved and cannot "leak away into the vacuum" is due to the invariance of the QED action under gauge transformations (A.6.2). More generally, symmetries (invariances) of the action always result in conserved quantities. Since 1949, unusual names have been given to similar conversed quantities: isospin, strangeness, color, charm, etc. The association of a conserved quantity with a differential symmetry of the action is known as Noether's Theorem, in honor of Emmy Noether who first showed this connection in 1915. A.7 Finding the gauge function Λ for the Lorentz Gauge: time-domain propagators In Fact 4 is was noted that if one already has a potential set ( A,φ), it is possible to find a new potential set (A',φ') such that div A' = f for any reasonable f. The method of finding the new set ( A',φ') was to find the function Λ from f as shown in (A.2.4) and then use the gauge transformation implied by Λ as shown in (A.4.1) to find the new potentials ( A',φ'). In the discussion of the Lorenz Gauge, we thus imagine we have some ( A,φ) and we want then to find a potential set ( A',φ') such that div A' = - 1 c2 ∂tφ', which is the Lorenz Gauge (A.6.3). We are thus using f = - 1 c2 ∂tφ' where φ ' is the partner to A'. One might fairly inquire what this function f actually is in terms of the starting potentials ( A,φ), since one does not a priori know what φ' is. In other words, since we don't a priori know what f(x) is, we cannot use (A.2.3) to find the right gauge function Λ to give the right new potentials ( A',φ'), so we seem to be in a circular conundr um when we try to fit this Lorentz gauge situation into the framework of our accumulated Facts above. Here is one way to find the right function Λ in terms of ( A,φ). We know from (A.2.3) and (A.4.1) that ∇ 2Λ = f = - 1 c2 ∂tφ' = - 1 c2 ∂t [φ - ∂tΛ ] . This can be written as follows, where the left side is the 3D wave equation operator acting on Λ, (∂t2 - c2∇2)Λ = ∂tφ . ( A . 7 . 1 ) Since we know φ from ( A,φ), we can obtain Λ by solving this differential equation. The equation is similar to the Poisson equation (A.0.1) when written this way in terms of the … symbol introduced above, c2… Λ = ∂tφ . // Stakgold (5.141) with u →Λ and q→∂ tφ (A.7.2) Here and below we include some supporting equatio n numbers from Stakgold Vol II. The formal solution of (A.7.2) can be found by first defining a Green's Function as we did above in (A.0.3), Appendix A: Gauge Invariance 230 c2… g(x,t; x',t') = δ (x-x')δ(t-t') // Stakgold (5.142) (A.7.3) The solution Green's function (propagator) is given by ← c 2 g(x,t; x',t') = (1/4 πR)δ(t-t'-R/c) with R = | x-x'| // Stakgold (5.155) n=3 (A.7.4) We have added an arrow that shows the direction of the propagator: it runs from time t' in the past to time t in the future, in which case t > t'. The propagator vanishes for all t < t' since in that case t-t'-R/c < 0 and the δ function can never get a hit. This g is an exam ple of a "causal" Green's function and it describes an expanding spherical wavefront seen at observation point x at time t propagating at velocity c from a point source at location x ' and time t' in the past. Formally one can then express a solution to (A.7.2) in a form similar to (A.0.2), Λ(x,t) = ∫d3x' ∫dt' g( x,t; x',t') ∂t'φ(x',t') . (A.7.5) Application of c2… to both sides of (A.7.5) with use of (A.7 .3) reproduces (A.7.2) showing that (A.7.5) is indeed the particular solution of (A.7.2). Inserting the propagator (A.7.4) we find that Λ(x,t) = 1 4πc2 ∫d3x' ∂tφ(x', t-R/c) R R = | x - x'| . (A.7.6) Thus we have solved our conundrum in that we have Λ expressed in terms of φ from the set ( A,φ). The solution (A.7.6) has the same form as the retarded solutions of Section 1.4. Once we have this Λ, we may use (A.4.1) to find the set ( A',φ') given the set ( A,φ). Comments: 1. Whereas the Poisson equation with ∇2 is "elliptic" in nature, the wave equation is "hyperbolic" since the various second derivatives in … don't all have the same sign, resulting in a change in the nature of the Green's function solution, the principle fact being that it is a causal function in terms of the time coordinates. For details on the above discussion, see Stakgold Vol II p 61-63 (fundamental solutions) and p 246-256 (Green's functions for the wave equation). Stakgold treats this subject with an arbitrary number of spatial dimensions n. One finds, for example, that for n = 3 the propagator (A.7.4) is an expanding infinitely thin spherical shell with no wake, whereas for n = 2 there is a wake behind the front as in his (5.151) which says g(r,t) = θ (t-r/v) 1/ (vt)2- r2 ( A . 7 . 7 ) where v is wave velocity. It is difficult to create a clea n unit impulse in water, but here is the rough idea: Appendix A: Gauge Invariance 231 http://physicsilluminati.blogs pot.com/2012/10/wa ve-optics.html 2. The time-domain Green's functions quoted in (A.7.4 ) for n = 3 and (A.7.7) for n = 2 are propagators for the wave equation (A.7.3) in 3D and 2D. When the se Green's functions are Fourier transformed to the frequency ω domain, they become the 3D and 2D Helmholtz propagators discussed in Appendix H and I, namely g F(r,r'; ω) = e-jkR/4πR = the Helmholtz 3D free-space propagator R = | r - r'| (H.1.7) g F(r,r'; ω) = (j/4) H 0(1)(kR) = the Helmholtz 2D free-space propagator k2 = ω2με (I.1.7) Appendix B: Magnetization Surface Currents on a Conductor 232 Appendix B: Magnetization Surface Currents on a Conductor Overview When a conductor of magnetic permeability μ 2 is embedded in a medium of μ 1 with μ1 ≠ μ2, a "bound current" (magnetization current) a ppears on the conductor surface. Section B.1 shows how this surface current K is related to the H field at the surface. Section B.2 shows how to compute H from the volume conduction current density Jc. Section B.3 then outlines a general plan for compu ting surface current K for an arbitrary conductor. Section B.4 computes H and the surface current for a round wire using symmetry. Section B.5 repeats the calculation using the general method outlined in Section B.3. Section B.6 presents what we call "the J m Theorem" which shows that adding the magnetization surface current of Section B.3 to the conduction cu rrent of a transmission line conductor adjusts the Helmholtz integral for A z so it gives the correct A z when the conductor and dielectric have different permeabilities, μ1 ≠ μ2. Section B.7 shows how this "J m Theorem" works for a round conductor. Plots are displayed for the three quantities A z, Bθ and Hθ. The conductors considered here are those of a transmission line in the "transmission line limit" in which it is assumed that the wavelength along the line is much l onger than the transverse dimensions of the line. In this case, it is reasonable to use 2D wave equations whose solutions then involve use of the 2D Poisson free-space propagator ln(R/2 π) as discussed in Appendix I. B.1 Relationship between surface current K and the field H at a conductor boundary First, consider this blowup of a piece of the boundary between a conductor (medium 2) and a dielectric (medium 1). Both media extend uniformly in the z direction, so we are looking at a piece of the cross section of a transmission line at a particular point on the surface of one of the conductors. Fig B.1 Appendix B: Magnetization Surface Currents on a Conductor 233 We shall assume that the conduction current is positive in the z direction, so J = Jzz^ with Jz > 0. Since the lower medium is the conductor in the drawing, the B and H field at the boundary are in the - x^ direction, that is to say, they point to the left due to the right hand rule relating J and B or H . According to (1.1.44), the tangential component of the H field is continuous at a boundary provided the boundary does not carry a free surface current, which is our situation here. Therefore, H x2 = Hx1 (1/ μ1)Bx1 = (1/μ2)Bx2 . (B.1.1) Assuming μ 2 ≥ μ1 (which would be the case if μ1 = μ0), the right equation implies |B x2| ≥ |Bx1| so the B field is larger inside the conductor. But in our picture, both B x2 and Bx1 are negative, so -B x2 ≥ - Bx1 which then says B x2 ≤ Bx1 and finally (B x2 - Bx1) ≤ 0. Also, H x2 = Hx1 ≤ 0. For the red loop shown in the figure one then has, as s → 0, ∫{ B • ds = Bx2L - Bx1L = (Bx2 - Bx1)L ≤ 0 . (B.1.2) Now consider (1.1.31) and (1.1.24) which say ( in the ω domain), ∫{ B • ds = ∫S curl B • dS = μ0 ∫S [ jωεE + Jc + Jm] • dS (B.1.3) where d S = dS z^. Since E • dS involves only E z (parallel to surface), and since by (1.1.41) such E z is continuous at the boundary, and since E z ≈ 0 inside the conductor, the ε jωE term makes no contribution, giving then ∫{ B • ds = μ0 ∫S [Jc + Jm] • dS . ( B . 1 . 4 ) As the distance s is taken to 0 in the red math loop above, ∫S Jc • dS → 0 because the conduction current is non-singular at the boundary. That is to say, ∫S Jc • dS → Jc • ∫S dS → 0. Since we shall take this limit in the end, we can then ignore the Jc term in (B.1.4) and write ∫{ B • ds = μ0 ∫S Jm • dS. intending to take s → 0 . (B.1.5) Since Jc flows in the + z^ direction, the surface current Jm flows in the - z^ direction (as shown below), so write Jm = Kz δ(y) z^ ( B . 1 . 6 ) where K z ≤ 0 is the magnitude of the surface current. Then Appendix B: Magnetization Surface Currents on a Conductor 234 ∫S Jm • dS = Kz ∫0 L dx ∫-s s dy δ(y) z^ • z^ = Kz ∫0 L dx = KzL . (B.1.7) Thus from (B.1.2), (B.1.5) and (B.1.7) we find that μ 0 Kz = Bx2 - Bx1 . ( B . 1 . 8 ) Compare this with (1.1.44) which says (as noted earlier, K zfree = 0 on our boundary) Kzfree = Hx2 - Hx1 . (1.1.44) The H field does not "see" our magnetization surface current K z, but the B field does see it. The signs are consistent with K z ≤ 0 and (B x2 - Bx1) ≤ 0 as noted above. Then from (B.1.1) we find μ 0Kz = Bx2 - Bx1 = (μ2Hx2 - μ1Hx1) = (μ2-μ1) Hx2 and finally K z = ( μ2 μ0 - μ1 μ0 ) Hx2 . ( B . 1 . 9 ) As noted earlier, H x2 < 0 so K z ≤ 0 is consistent with μ2 ≥ μ1. We now rewrite this result in terms of a different picture: Fig B.2 This shows the cross section of the entire conductor in gray, and Jc is still directed toward the viewer. In this picture a point on the surface is associated with a local coordinate system for which r^ = y^ is normal to the surface and θ^ = -x^ is tangent to the surface (so H θ = -Hx). We are thinking of (r, θ,z) as local cylindrical coordinates at the point shown on the conductor surface, where r^ x θ^ = z^, and the x,y,z directions of the figure match those of the previous figure where as usual x^ x y^ = z^ . Then (B.1.9) says Appendix B: Magnetization Surface Currents on a Conductor 235 Kz = - ( μ2 μ0 - μ1 μ0 ) Hθ a m p s / m ( B . 1 . 1 0 ) where H θ > 0 and K z ≤ 0. In general, K z is a function of position on the perimeter of the conductor cross section. We can compute the total magnetization current I m (amps) by integrating K z around the perimeter of the conductor: ∫C Kz ds = - ( μ2 μ0 - μ1 μ0 ) ∫C Hθ ds = - ( μ2 μ0 - μ1 μ0 ) ∫{C H • ds But by (1.1.37), ∫{C H • ds = ∫S [jωεE+J] • dS = ∫S [jωεEz+Jz] dS ≈ ∫S Jz dS = I and therefore Im ≡ ∫C Kz ds = - ( μ2 μ0 - μ1 μ0 ) I ( B . 1 . 1 1 ) The ratio of the magnetization current to the conduction current is given by constant f m , fm ≡ Im/ I = - ( μ2 μ0 - μ1 μ0 ) . ( B . 1 . 1 2 ) and this result is independent of the shape of the conductor. Of course if μ1 = μ2, there is no magnetization current and K z and Im are both zero. Example : For a round wire of radius a carrying an axially symmetric current distribution, we know that 2πaHθ = I so H θ = I/(2πa) at the surface. Then Kz = - ( μ2 μ0 - μ1 μ0 ) [ I/(2πa)] . // round wire of radius a and μ2, dielectric μ1 (B.1.13) The total surface magnetization current integrat ed around the round wire surface is then Im = 2πaKz = - ( μ2 μ0 - μ1 μ0 ) I ( B . 1 . 1 4 ) in agreement with (B.1.11). For example, if the dielectric has μ1 = μ0 and the conductor has μ 2 = 2μ0, then Imag = - I. We return to this example in Section B.4 below. Appendix B: Magnetization Surface Currents on a Conductor 236 Physical mechanism of the magnetization surface current. As a reminder, a surface magnetization current arises at a boundary between media with different μ values just the way surface polarization charge arises at a boundary between media with different ε. In the μ case, here is a suggestive picture : Fig B.3 On the left we look at a round wire end on, while the right shows a top view where the front end of the wire on the left has been tilted down. Here μ 1= μ0 so there is only vacuum outside the wire. The B field lines up the little magnetic dipoles (or creates them) accord ing to the right hand rule which we represent schematically as little atoms with orbiting electrons. On the right, B comes out of paper and lines up the magnetic moments CCW as shown there. On the left , B goes into paper so the moments are lined up clockwise instead. In both cases, the resulting magnetiza tion surface current is in the same direction, as indicated by the arrows of the loops hanging outside the wire. The picture shows why it is that the surface current is directed opposite to the current J which creates it, a sort of magnetic Lenz's Law. Note that this surface current is "not seen" by H, but it is seen by B, as mentioned in (1.1.24). In the case that the outer medium has some μ1 > μ0, both media have surface currents at the boundary, and then when μ1 ≠ μ2 there is a surface current imbalance resulting in a net surface current. If it happens that μ1 < μ2, then the directions shown above are correct, but if μ1 > μ2, the surface current runs in the opposite direction to that shown. There is also a bulk volume magnetization current away from the surface, not shown above. Details of the magnetization current J m for a round wire are computed (DC) in Appendix G (G.3.4) where the current density J mz includes a surface delta function. B.2 Calculation of H from the current J in a conductor (a) An expression for H in terms of J Start with Maxwell's equation (1. 1.1), and we are now working inside a conductor so E = 0 and then curl H = jωεE + J = J ( B . 2 . 1 ) where J is the conduction current. Apply curl to both sides and use curl curl = grad div - ∇ 2 to get grad div H - ∇ 2H = curl J . (B.2.2) Appendix B: Magnetization Surface Currents on a Conductor 237 But in a uniform medium div H = 0 since div B = 0 so ∇2H = - curl J . ( B . 2 . 3 ) Now let's assume that we have J = Jz(x,y) z^ and assume that the solution H does not depend on z. In that case we have ∇2 2D H(x,y ) = - curl J J = Jz(x,y) z^ . (B.2.4) The particular solution to this PDE is shown in (I.1.8) to be H(x,y) = ∫d2x' [ 1 2π ln(1/R) ] curl' J (x') R = | x-x'| (B.2.5) or H(x,y) = - 1 4π ∫d2x' ln(R2) curl' J(x') R = | x-x'| (B.2.6) where 1 2π ln(1/R) is the Poisson 2D free-space propagator. This gives H in terms of J . (b) An alternative derivation using the vector potential A z An alternate derivation of (B.2.5) makes use of the vector potential A z. Start with (1.5.4), ( ∇2 + β2)A(x) = - Σi μiJi (x) . all of region R (1.5.4) Apply this to a single conductor and assume A(x) = Az(x,y) z^, so the above equation becomes - ∇2 2D Az(x) = μJz(x) . ( B . 2 . 7 ) The particular integral from (I.1.8) is then A z(x) = ∫d2x' [ln(1/R)/2 π] [μJz(x') = - (μ/2π) ∫d2x' ln(R) J z(x') . (B.2.8) Then use B = curl A = x^ (∂ yAz - ∂zAy) + y^ (∂zAx - ∂xAz) + z^ (∂xAy - ∂yAx) = x^ (∂ yAz) + y^ (- ∂xAz) ( B . 2 . 9 ) so that μH = x^ (∂ yAz) + y^ (- ∂xAz) . ( B . 2 . 1 0 ) Appendix B: Magnetization Surface Currents on a Conductor 238 Now compute, ∂yAz(x) = ∂y [- (μ/2π) ∫d2x' ln(R) J z(x') ] = - (μ/2π) ∫d2x' Jz(x') ∂y ln(R) . But ∂y ln(R) = - ∂ y' ln(R) since R = | x-x'|. But then do parts integration to move ∂y'to Jz(x') picking up an offsetting minus sign, and the parts vanish on a great circle surrounding the conductor. Thus, ∂yAz(x) = - (μ/2π) ∫d2x' ln(R) ∂y'Jz(x') ∂xAz(x) = - (μ/2π) ∫d2x' ln(R) ∂x'Jz(x') . (B.2.11) Then from (B.2.10) one gets, H = - (1/2π)∫d2x' ln(R) [ x^ ∂y'Jz(x') - y^ ∂x'Jz(x')] . But the coordinates x = (x,y) and x' = (x',y') have the same unit vectors x^ = x^' and y^ = y^' . Since J = Jzz^ we then end up with H(x) = - (1/2π)∫d2x' ln(R) curl' J z(x') = + (1/2 π)∫d2x' ln(1/R) curl' J(x') which agrees with (B.2.5). (c) Boundary conditions Recall from (B.1.1) that the tangential component of H is continuous through the boundary between conductor and dielectric, even if μ1 ≠ μ2. Consider then the transverse component of (B.2.5) at some point on the conductor surface such as the point shown in Fig 3.3. We have (t = transverse) Ht(x) = ∫d2x' [ 1 2π ln(1/R) ] [curl' J(x')]t R = | x-x'| , (B.2.12) This particular integral is natura lly continuous at the boundary between the media, and this agrees with the fact that H t(x) must have this property. Therefore, no homogeneous solutions of (B.2.4) ∇2 2DHt = 0 need be added in, so (B.2.12) is the complete solution for H t(x). This solution can then be used in (B.1.10) to find the magnetization surface current. Appendix B: Magnetization Surface Currents on a Conductor 239 (d) The Biot-Savart Law in 3D and 2D Recall the ab ove 3D vector Helmholtz equation, ∇2H = - curl J . (B.2.3) In Cartesian coordinates it is three scalar Helmholtz eq uations which can be solved as in (H.1.8) to give H(x) = ∫d3x' [ 1 4πR ] curl' J(x') R ≡ |x-x'| ( B . 2 . 1 3 ) where 1/4 πR is the 3D free-space Poisson propagator. Write this in components and define R as shown, Hi(x) = ∫d3x' [ 1 4πR ]εijk ∂'jJk(x') , R ≡ x - x' = points to observation point x . (B.2.14) Then move ∂j' from Jk to (1/R) by parts integration (pick up minus sign) and throw out the parts for the usual reasons (see end of Appendix A.1), Hi(x) = - ∫d3x' 1 4π ∂'j (1 R ) εijk Jk(x') . (B.2.15) Then note that ∂'jR-1 = -R-2 ∂'jR and ∂'jR = ∂'j Σk(x'k-xk)2 = (1/2)(1/R) 2(x' j-xj) = R-1 (x'j-xj) = - R-1 Rj (B.2.16) so that ∂'jR-1 = +R-3Rj. Only the parts minus sign remains, so Hi(x) = - ∫d3x' [ 1 4πR3 ]εijk Rj Jk(x' ) ( B . 2 . 1 7 ) or reversing the cross product order, H(x) = ∫d3x' 1 4πR3 J(x') x R . R ≡ x - x' ( B . 2 . 1 8 ) This equation is basically the 3D Biot-Savart Law , see for example Panofsky and Philips p 125 (7.31). For a short piece d s' of thin wire carrying current I, one writes J(x') d3x' = I d s' so the above becomes, H(x) = ∫{ 1 4πR3 I ds' x R or d H(x) = 1 4πR3 I ds' x R . (B.2.19) We can apply the same process to obtain a 2D Biot-Savart Law as follows. Start with Appendix B: Magnetization Surface Currents on a Conductor 240 ∇2 2D H(x,y ) = - curl J J = Jz(x,y) z^ (B.2.4) and its solution (B.2.5) H(x,y) = ∫d2x' [ 1 2π ln(1/R) ] curl' J (x') R = | x-x' | (B.2.5) where 1 2π ln(1/R) is the Poisson 2D free-space propagator. Then, inverting 1/R, H i(x,y) = - ∫d2x' [ 1 2π ln(R) ] εijk ∂'jJk(x') . (B.2.20) Doing the same parts integration gives Hi(x,y) = + ∫d2x' [ 1 2π ∂'j ln(R) ] ε ijk Jk(x' ) ( B . 2 . 2 1 ) and now using result (B.2.16) from above, ∂'j ln(R) = R-1∂'jR = R-1 [- R-1 Rj] = -R-2Rj ( B . 2 . 2 2 ) we get Hi(x,y) = - ∫d2x' [ 1 2π R-2Rj ] εijk Jk(x') = - ∫d2x' 1 2πR2 εijk Rj Jk(x') (B.2.23) or H(x,y) = ∫d2x' 1 2πR2 J(x') x R R ≡ x - x' ( B . 2 . 2 4 ) which is the 2D Biot-Savart Law . It provides an alternate way to obtain H from J in a 2D problem. B.3 General Method for computing the surface current J m on a wire Here are the steps for a wire of ar bitrary cross sectional shape: 1. Compute the H field at all points in the wire cross-sectional plane section using (B.2.6) or (B.2.24) H(x,y) = - 1 4π ∫d2x' ln(R2) curl' J(x') R = | x-x' | . (B.2.6) H(x,y) = ∫d2x' 1 2πR2 J(x') x R R ≡ x - x' (B.2.24) or use the third method of first computing A z, Appendix B: Magnetization Surface Currents on a Conductor 241 Az(x) = - (μ/2π) ∫d2x' ln(R) J z(x' ) (B.2.8) B = curl A = x^ (∂yAz) + y^ (- ∂xAz) H = B/μ (B.2.9) 2. Evaluate this H field at xb = (xb,yb) for all points x b on the cross section boundary. 3. Compute the component of H which is tangential to the boundary in the cross sectional plane. Call this component H θ. 4. The surface current density is then given by (B.1.10), K z = - ( μ2 μ0 - μ1 μ0 ) Hθ a m p s / m (B.1.10) B.4 Surface current on a round wire with uniform J For a round wire of radius a with uniform J z (as would be the DC case ω = 0), geometric symmetry makes the calculation of H very easy. One need only apply Ampere's Law separately for a point r outside the wire, and for another point r inside the wire. For the outside case one finds 2πr Hθ(r) = I => H θ(r) = I/(2πr) r ≥ a . (B.4.1) And then for the inside case the "current enclosed" is determined by a simple area fraction. 2πr H θ(r) = I (πr2/πa2) => H θ(r) = I r/(2 πa2) r ≤ a . (B.4.2) At the boundary the two expressions agree and we have H θ = I/(2πa ) . ( B . 4 . 3 ) If this wire has magnetic permeability μ 2 and is embedded in an infinite medium of μ1, then the surface magnetization current induced on the wire is Kz = - ( μ2 μ0 - μ1 μ0 ) Hθ = - ( μ2 μ0 - μ1 μ0 ) I/(2πa) amp/m (B.4.4) Kz = Kzz^ ( B . 4 . 5 ) and this surface current is in the direction opposite J if μ2 > μ1. If μ1 = μ2, the surface current vanishes. This surface current could be expressed in volume density form as Jm = Kzδ(r-a)z^ amp/m2 . ( B . 4 . 6 ) Appendix B: Magnetization Surface Currents on a Conductor 242 B.5 Computing H θ for a round wire using the General Method of B.3 For a wire of some general cross section, symmetry is not available to allow the simple solution for H θ outlined in the previous section. We then have to use the more general method outlined in Section B.3 above. As a check on the viability of this general met hod, we shall carry out "step 1" of the method and show how H θ may be computed from J using the formula (B.2.6). The conduction current density in a round wire with uniform J z is given by J z(r) = J0θ(a-r) (B.5.1) where θ is the Heaviside step function. Our first step is to compute curl J, and we do this in cylindrical coordinates by just staring at the cylindrical-coordinates curl formula, curl J = r^ [ r -1∂θJz - ∂zJθ] + θ^ [∂zJr - ∂rJz] + z^ [ r-1∂r(rJθ) - r-1∂θJr ] (B.5.2) and finding the only non-zero piece which is this (uniform J z), curl J = [-∂rJz(r)] θ^ . ( B . 5 . 3 ) Inserting J z(r) from above we find ∂r Jz(r) = J0 ∂rθ(a-r) = - J 0 δ(r-a) (B.5.4) => curl J(r) = θ^ J0 δ(r-a) . (B.5.5) so we have a "ring source of curl J". For use in our integral for H we then have curl' J(r') = θ^' J0 δ( r ' - a ) . ( B . 5 . 6 ) For a current distribution which tapers off smoothly to 0 at the wire edge one w ould not have this singular contribution, but for a wire with prescribed uniform current, it is present, and curl J vanishes everywhere but on the boundary. The relevant picture is this: Appendix B: Magnetization Surface Currents on a Conductor 243 Fig B.4 From (B.2.5) the H field at any point x = (x,y) is then given by H(x,y) = - 1 4π ∫d2x' ln(R2) curl' J (x') = - 1 4π ∫d2x' ln(R2) θ^' J0 δ(r'-a) = - J0a 4π ∫-π π dθ' ln(R2)|r'=a θ^' or H(r,θ) = - J0a 4π ∫-π π dθ' ln [ r2 + a2 - 2ar cos( θ'-θ) ] θ^' . (B.5.7) The figure shows that θ^' = cosθ' y^ - sinθ' x^ ( B . 5 . 8 ) so then H(r,θ) = - J0a 4π ∫-π π dθ' ln [ r2 + a2 - 2ar cos( θ'-θ) ] [cosθ' y^ - sinθ' x^] . (B.5.9) Next, let x ≡ θ'-θ. Since the ∫dθ' has full range 2 π, one can replace ∫-π π dθ' = ∫-π π dx . Then H(r,θ) = - J0a 4π ∫-π π dx ln [ r2 + a2 - 2ar cos(x) ] [cos(x+θ) y^ - sin(x+θ) x^] . (B.5.10) Now writing H = Hx x^ + Hy y^ , decompose the above into two equations Hx(r,θ) = + J0a 4π ∫-π π dx ln [ r2 + a2 - 2ar cos(x) ] sin(x+ θ) Hy(r,θ) = - J0a 4π ∫-π π dx ln [ r2 + a2 - 2ar cos(x) ] cos(x+ θ) (B.5.11) or Appendix B: Magnetization Surface Currents on a Conductor 244 Hx(r,θ) = + J0a 4π ∫-π π dx ln [ r2 + a2 - 2ar cos(x) ] [ sinxcos θ+cosxsinθ ] Hy(r,θ) = - J0a 4π ∫-π π dx ln [ r2 + a2 - 2ar cos(x) ] [cosxcos θ - sinxsinθ ] . (B.5.12) Since ∫-π π dx is over an even range, throw out odd integrand terms, and then fold the negative range into the positive adding a factor of 2 to get Hx(r,θ) = + sinθ J0a 2π ∫0 π dx ln [ r2 + a2 - 2ar cos(x) ] cosx ≡ sinθ J0a 2π Q Hy(r,θ) = - cosθ J0a 2π ∫0 π dx ln [ r2 + a2 - 2ar cos(x) ] cosx = -cos θ J0a 2π Q (B.5.13) where Q ≡ ∫0 π dx ln [ r2 + a2 - 2ar cos(x) ] cosx . (B.5.14) Before evaluating this integral, we see that H = H x x^ + Hy y^ = - J0a 2π Q [ cosθ y^ -sinθ x^ ] = - J0a 2π Q θ^ = Hθ θ^ . (B.5.15) Thus we find that the resulting H is entirely in the θ^ direction and Hθ = - J0a 2π Q . ( B . 5 . 1 6 ) We seek now to evaluate this integral Q, Q ≡ ∫0 π dx ln [ r2 + a2 - 2ar cos(x) ] cosx = ∫0 π dx ln [ {a2}{ (r/a)2 + 1 - 2(r/a) cos(x)} ] cosx = ∫0 π dx { ln (a2) + ln[(r/a)2 + 1 - 2(r/a) cos(x)] } cosx = l n ( a2)[ ∫0 π dx cosx ] + ∫0 π dx ln[(r/a)2 + 1 - 2(r/a) cos(x)] cosx = l n ( a2)[0] + ∫0 π dx ln[α2 + 1 - 2α cos(x)] cosx where α ≡ r/a Appendix B: Magnetization Surface Currents on a Conductor 245 = ∫0 π dx ln[α2 + 1 - 2α cos(x)] cosx . // = Q (B.5.17) This integral is the n=1 special case of the following integral from GR7 page 589 4.379.6, Therefore we find Q = ⎩⎨⎧ -π(r/a) r<a -π(a/r) r >a ( B . 5 . 1 8 ) so Hθ = - J0a 2π Q = ⎩⎨⎧ (aJ0/2) (r/a) r<a (aJ0/2) (a/r) r >a . (B.5.19) Now the total current in the wire is I = J 0πa2 so (aJ0/2) = (I/2πa) and then Hθ = ⎩⎨⎧ (I/2πa) (r/a) r<a (I/2πa) (a/r) r >a = ⎩⎨⎧ I (r/2πa2) r<a I (1/2πr) r >a . (B.5.20) Thus, we finally arrive at the same results for H θ as obtained in (B.4.2) and (B.4.1). B.6 Modification of King's Helmholtz integral solution when μ1 ≠ μ2 (a) General Discussion In Section 4. 7 we wrote the vector potential for transmission line conductor C in this manner, Az(x,ω) = 1 4π ∫ C μ2 Jzc(x',y',z',ω) e-jβdR R dx'dy'dz' . R = | x - x'| (4.7.2) where we have made a notational cha nge to be consiste nt with previous sections of Appendix B. Here we shall use μ 2 to refer to the permeability of the conductor, and μ1 to be that of the dielectric (these are called μ and μd in Section 4). The above "Helmholtz integr al" is only the "particular solution" of the Helmholtz equation ( ∇2 + β12)Az= -μ2Jcz. When μ 1 ≠ μ2, it turns out that one must add a homogeneous solution A z(homo) [ that is, ( ∇2 + β12) Az(homo) = 0] to the Helmholtz solution shown above in order to meet boundary conditions. Appendix G.4 provides a very detailed study of just how this works for a round conductor with a uniform current distribution. Appendix B: Magnetization Surface Currents on a Conductor 246 To avoid this major complication, we limite d the analysis of Chapter 4 to the case that μ1 = μ2. This means, for example, that Chapters 4,5,6 are applicable for non-magnetic c onductors in a non-magnetic dielectric, in which case μ1 = μ2 = μ0. The work presented below generalizing to μ1 ≠ μ2 is then summarized in Section 4.12. With the reader's permission, we replicate th e following comments from below (4.7.6), making a few small changes: Comments regarding μ This is a subtle subject and is not discus sed in King's transmission line theory book. If the dielectric and conductor have the same permeability so that μ 1 = μ2, then there exists no "magnetic boundary" between the conducto r and dielectric. The solution (4.7.2) is then smooth at this boundary, and so A z(x,y,z) "naturally" satisfies these two boundary conditions, A z(x+) = Az(x- ) (1/μ1) ∂nAz(x+) = (1/μ 2) ∂nAz(x-) (4.7.7) (B.6.0) where x+ is just outside the conductor surface and x- is just inside. The second equation here is just (1.1.46) in the case that there is no free surface current K free flowing on the boundary, and indeed in our example at hand there is no such free surface current. Since we have assumed that μ1 = μ2, this second boundary condition just says ∂ nAz(x+) = ∂nAz(x-). Since there is no magnetic boundary at the conductor/dielectric interface, the so lution (4.7.2) is continuous and all its derivatives are also continuous at the boundary, since nothing special happens at that boundary. Thus, the Helmholtz integral solution provides the whole solution for A z since it automatically meets both "boundary conditions" at this pseudo boundary. If on the other hand we have μ1 ≠ μ2, then there is a magnetic boundary between conductor and dielectric which we have to worry about. In this case, (4.7.2) cannot possibly satisfy the second boundary condition of (B.6.0) since, as already noted, the A z of (4.7.2) satisfies ∂nAz(x+) = ∂nAz(x+). Thus, in this case (4.7.2) is not the full solution for A z. One must add a homogeneous He lmholtz equation solution to (4.7.2) in order to have a proper solution for A z that satisfies both equations in (B.6.0). It turns out that the correct total A z solution can be generated by adding a certain fictitious surface current to μ2Jz in (4.7.2). Since such a surface current vanishes on both sides of the boundary between μ1 and μ2, the Helmholtz solution due just to this surface current is in fact a homogeneous solution to the Helmholtz equation in both the conductor and dielectric regions, away from that boundary. It turns out moreover that the correct fictitious surface current to add is in fact the magnetization surface current J m which is created at the boundary between μ1 ≠ μ2. Adding this surface current is just a "trick" in order to generate the correct homogeneous adder solution so that the resulting total A z satisfies both boundary conditions in (B.6.0). Formally speaking, the J appearing in (1.5.3) and then J z in (4.7.2) should not include such magnetization currents since this J is really the J in Maxwell's equation curl H = ∂tD + J, and this J does not include magnetization currents -- it includes only normal conduction currents. Here we wish to prove the claim that adding the surface magnetization current to the conduction current does in fact make the boundary conditions work. After doing this proof, we will show in Section B.7 just how this works out in the case of a round conductor. Appendix B: Magnetization Surface Currents on a Conductor 247 We stress that only the surface part of J m gets added in. In general J m will also have a "bulk" component in the dielectric and conductor. If we were to include this bulk component, we would not be adding a homogeneous solution to the particular solution, and we would in fact be creating a non-solution! In (G.3.4) we show the complete Jm for a round wire carrying a uniform current, and it does have both bulk and surface components. (b) Statement and Proof of the J m Lemma The Jm Lemma. If Jmz represents the surface component of magnetization current density for a transmission line conductor C of μ2 with conduction current density J cz, embedded in a dielectric medium of μ1, then if we write Az(x) = 1 4π ∫ C [μ2 Jcz(x') + μ0 Jmz(x')] e-jβdR R dV' . R = | x - x'| (B.6.1) this Az(x) will satisfy the boundary co nditions (B.6.0) shown above. Comments. 1. In this Lemma, we are showing that the contribution to A z(x) just from conductor C satisfies the boundary condition (B.6.0) at the surface of conducto r C if we add in the surface current term μ 0 Jmz(x') as shown. What we really want to show is that the total Az(x) due to all the transmission line conductors satisfies (B.6.0) at the surface of conductor C. This will be the content of the J m Theorem presented in Section (c) below. Once this Theorem is proved, we know that "the other conduc tors" don't interfere with the Jm Lemma and we can regard the boundary conditions on A z(x) given in (B.6.1) as applying also to the full A z(x) which includes the contributions of all conductors. 2. Our proof below applies in the "transmission line limit" of Sections 4.3 and 4.9 which is essentially a long wavelength and small β limit. In this limit, we can replace our various Helmholtz propagators below with Poisson (Laplace) propagators. Nevertheless, we maintain the Helmholtz forms in the hope that the above theorem is valid for reas onably moderate (but not huge) β values. At very large β values the whole transmission line framework collapses anyway, transverse A x and Ay components are no longer small, and the line picks up transverse waveguide activity. Appendix B: Magnetization Surface Currents on a Conductor 248 (1) Preliminaries We first quote a key result from Stakgold concerning boundary layer a( ξ): u(x) = ∫σ dSξ a(ξ) E(x|ξ) u(s) = ∫σ dSξ a(ξ) E(s|ξ) ( B . 6 . 2 ) ∂ νu(x) = ∫σ dSξ a(ξ) ∂νE(x|ξ) ∂νu(s) = [∂nu(x)]x→s± = ∫σ dSξ a(ξ) ∂νE(s|ξ) ∓ a(s)/2 // extra term ! (B.6.3) This is a tricky subject and some words are certainly in order. In the Stakgold world, σ is a surface of n-1 dimensions existing in an n dimensional space. E(x| ξ) is the free-space propagator in that n dimensional space (the "fundamental solution"). The in tegrals shown above are over the surface σ, and ξ represents the n-1 dimensional coordinate of a point on the surface σ, while dS ξ is a piece of "area" on the surface. (Stakgold does not write vectors in bold font as we do in this document.) Function a( ξ) is defined on the surface and is called a simple (monopole) boundary layer. Stakgold also deals with dipole layers (as in a cell membrane), but we don't care about them right now. The question at hand is this: What happens as a point x away from the surface approaches the surface where it becomes point s? We are interested in the limit x → s. As shown in the first pair of equations (B.6.2), nothing unusual happens for the function u(x) defined as shown by the integral. One then says that u(x) is "continuous" at x = s. But so mething very unusual happens for the function ∂ νu(x) where ∂ν denotes a derivative locally normal to the surface at s. As x → s an "extra term" a ppears as shown above having value ∓ a(s)/2. If normal ν points "out" from the surface then as one approaches from the outside (call it the + side), the extra term is -a(s)/2, but if the approach is from the inside (- side), the extra term changes sign. Here is a picture illustrating the ge ometry of the above equations: ( n is normal at ξ , ν is normal at s) F i g B . 5 The reason the extra term appears ha s to do with the nature of the d ξ integration when ξ is very close to s which is somewhat of a singular situation since R ≡ |s-ξ| → 0. Stakgold treats surface layers in Section 6.4 of his Volume II, pages 110-120, and his treatment involves a lot of detail. The claims shown above Appendix B: Magnetization Surface Currents on a Conductor 249 appear on pages 118 and 119, though the conclusions ar e a bit obscured in the detail. Stakgold works in 3D with E(x| ξ) = (1/4π|x-ξ|) = 1/4π R and often uses these quantities, k( s,ξ) = cos( s -^ ξ, n^)/ [4π|s-ξ|2] // cos(upper marked angle) k(ξ,s) = cos(ξ -^ s, ν^)/[4π|s-ξ|2] = ∂νE(s|ξ) . // cos(lower marked angle) Later in his Problem 6.18 through 6.20 Stakgold has the r eader verify that the resu lts are also valid in 2D where surface σ is then just a curve. These are the results we shall use. Although he does not state it outright, we think his results are probably valid for σ being a surface of any number of dimensions, but our only interest will be the 2D case. (2) Outline of Proof of the J m Lemma We break up (B.6.1) into these two terms, a "conduction term" and a "magnetization" term, Az(c)(x) = 1 4π ∫ C [μ2 Jcz(x')] e-jβdR R dV' . R = | x - x'| (B.6.4) Az(m)(x) = 1 4π ∫ C [μ0 Jmz(x')] e-jβdR R dV' . R = | x - x'| (B.6.5) Az(x) = Az(c)(x) + Az(m)(x) . two terms (B.6.6) As noted earlier, the first term is smooth at a point s on a conductor surface and satisfies the two boundary conditions, Az(c)(s+) = Az(c)(s- ) ∂nAz(c)(s+) = ∂nAz(c)(s-) . (B.6.7) so one can write A z(c)(s) or ∂nAz(c)(s) without concern for whether s is s+ or s-. For this term, which is the Helmholtz "particular" integral, we may then trivially write, 1 μ1 ∂nAz(c)(s+) – 1 μ2 ∂nAz(c)(s-) = + [ 1 μ1 - 1 μ2 ] ∂nAz(c)(s) . (B.6.8) Since this is non-zero, the term A z(c) on its own does not meet the required slope boundary condition (B.6.0) at an interface between μ 1 and μ2, and that is precisely why we need the A z(m) term. Our goal is to show that 1 μ1 ∂nAz(m)(s+) – 1 μ2 ∂nAz(m)(s-) = – [ 1 μ1 - 1 μ2 ] ∂nAz(c)(s) (B.6.9) so that when we add the two terms we will get Appendix B: Magnetization Surface Currents on a Conductor 250 1 μ1 ∂nAz(s+) – 1 μ2 ∂nAz(s- ) = 0 ( B . 6 . 1 0 ) as required by (1.1.46). The concludes our proof outline , and it remains then to demonstrate (B.6.9). (3) Verification of (B.6.9) We start with (B.6.5) where ∫dV' is over the entire transmission line conductor C, Az(m)(x) = 1 4π ∫ C dV' [μ0 Jmz(x')] e-jβdR R = ∫ C dV' [μ0 Jmz(x') E3(x|x' ) ( B . 6 . 1 1 ) where E 3(x|x') = e-jβR 4πR → 1 4πR as β→0 . ( B . 6 . 1 2 ) We know from Chapter 4 that in the transmission line lim it we can do the dz' integration in dV' and arrive at a 2D-propagator expression for the above potential, where the integral is now over the cross section area of the conductor C, A z(m)(x) =∫ C dS' [μ0 Jmz(x')] E2(x|x' ) ( B . 6 . 1 3 ) where E 2(x|x') = (j/4) H 0(1)(kR) → - 1 2π ln(R2) as β→0 . (B.6.14) This whole subject of transitioning from the 3D to 2D an alysis is reviewed in A ppendix J, and it occurs in many places in this document. We are only using that portion of J mz which is a surface current on the perimeter of C, so we rewrite the above as Az(m)(x) = ∫{C ds' [μ0 Kz(x')] E2(x|x' ) ( B . 6 . 1 5 ) where K z(x') is the magnetization surface current (amps/m) discussed in Section B.1. Stakgold's surface integral over σ is now just a line integral around the perime ter of the conductor C cross section. Recall from (B.1.10) that the magnetization surface current is given by, K z = - ( μ2 μ0 - μ1 μ0 ) Hθ (B.1.10) Appendix B: Magnetization Surface Currents on a Conductor 251 where H θ is the H field tangent to the cross section surface. Inserting this K z into (B.6.15) gives Az(m)(x) = ∫{C ds' [(μ1-μ2) Hθ(x')] E2(x|x') . (B.6.16) We now identify this with the first of Stakgold' s equations (B.6.2) and we know we can take x→s with no surprises. If we now replace Stakgold's normal direction ν with our usual normal symbol n, we can write (B.6.3) as ∂nAz(m)(x) = ∫{C ds' [(μ1-μ2) Hθ(x')] ∂nE2(x|x' ) ( B . 6 . 1 7 ) ∂nAz(m)(s±) = ∫{C ds' [(μ1-μ2) Hθ(x')] ∂nE2(s|x') ∓ [(μ1-μ2) Hθ(s)]/2 . (B.6.18) Now since we want to prove (B.6.9), we first evaluate its left hand side using (B.6.18) twice, 1 μ 1 ∂nAz(m)(s+) – 1 μ2 ∂nAz(m)(s-) = 1 μ1 { ∫{C ds' [(μ1-μ2) Hθ(x')] ∂nE2(s|x') - [(μ1-μ2) Hθ(s)]/2 } – 1 μ2 { ∫{C ds' [(μ1-μ2) Hθ(x')] ∂nE2(s|x') + [(μ1-μ2) Hθ(s)]/2 } (B.6.19) = ( μ1-μ2) [ 1 μ1 - 1 μ2 ] ∫{C ds' Hθ(x') ∂nE2(s|x') – [ 1 μ1 + 1 μ2 ] (μ1-μ2) Hθ(s)/2 . Our task of showing that (B.6.9) is true then boils down to showing that the last expression above is equal to – [1 μ1 - 1 μ2 ] ∂nAz(c)(s) . That is to say, we have to show (μ1-μ2) [1 μ1 - 1 μ2 ] ∫{C ds' Hθ(x') ∂nE2(s|x') – [1 μ1 + 1 μ2 ] (μ1-μ2) Hθ(s)/2 = – [1 μ1 - 1 μ2 ] ∂nAz(c)(s) ? ( B . 6 . 2 0 ) A question mark indicates an equation that we want to show is true, but have not yet done so. Canceling ( μ1-μ2) factors, (B.6.20) becomes [1 μ1 - 1 μ2 ] ∫{C ds' Hθ(x') ∂nE2(s|x') – [1 μ1 + 1 μ2 ] Hθ(s)/2 = + 1 μ1μ2 ∂nAz(c)(s) ? (B.6.21) or (μ2-μ1) ∫{C ds' Hθ(x') ∂nE2(s|x') – (1/2)( μ2+μ1) Hθ(s) = ∂nAz(c)(s ) . ? (B.6.22) The integral in (B.6.22) can be replaced using (B.6.18) with the s+ choice, Appendix B: Magnetization Surface Currents on a Conductor 252 ∂nAz(m)(s+) = ∫{C ds' [(μ1-μ2) Hθ(x')] ∂nE2(s|x') - (1/2)[(μ 1-μ2) Hθ(s) (B.6.18)+ so (μ 2-μ1) ∫{C ds' Hθ(x') ∂nE2(s|x') = – ∂nAz(m)(s+) – (1/2) (μ1-μ2) Hθ(s) . (B.6.23) Equation (B.6.22) then becomes, – ∂ nAz(m)(s+) – ( 1/2)(μ1-μ2) Hθ(s) – (1/2)(μ 2+μ1) Hθ(s) = ∂nAz(c)(s) ? or – ∂ nAz(m)(s+) – μ1 Hθ(s) = ∂nAz(c)(s) ? or – μ 1 Hθ(s) = ∂n [Az(c)(s) + Az(m)(s+) ] ? or – μ 1 Hθ(s) = ∂nAz(s+) . ? // using (B.6.6) (B.6.24) We now introduce a local cylindrical coordina te system in this manner relative to point s Fig B.6 Notice that r^ is the normal vector at point s, so ∂r = ∂n. Then first we determine B θ, B = curl A = r^ [ r-1∂θAz - ∂zAθ] + θ^ [∂zAr - ∂rAz] + z^ [ r-1∂r(rAθ) - r-1∂θAr ] = r^ [ r-1∂θAz] + θ^ [- ∂rAz] = θ^ [- ∂rAz] . Here we have set ∂ θAz = 0 according to Fact 7 of (3.8.11) which says A z is constant on the cross section surface (which implies the strong or extreme sk in effect regime). The result is then, B θ(s+) = - ∂nAz( s + ) . ( B . 6 . 2 5 ) Appendix B: Magnetization Surface Currents on a Conductor 253 Since s+ is in the dielectric with μ1 we then have H θ(s+) = (1/μ 1) Bθ(s+) = - (1/ μ1) ∂rAz(s+) // H θ(s+) = H θ(s-) = H θ(s) says (1.1.42) so -μ1 Hθ(s) = ∂rAz(s+) = ∂nAz( s + ) . ( B . 6 . 2 6 ) But this last equation matches our equation in questio n (B.6.24), so we can then go back and erase all the question marks and we have then verified e quation (B.6.9) and our proof is complete. Comment : We noted that Stakgold's analysis is qu ite complicated. He uses the Laplace free-space propagators such as E 2(x|x') = -(1/2 π) ln(R2), but we think his analysis also applies for the Helmholtz propagators. The reason is that the Helmholtz complica tion does not really change the singular nature of things near R = 0. This is most obvious when comparing e-jβR/4πR to 1/4πR. If we are wrong about this conjecture, we can regard the above theorem as proven only for small β which in fact defines the transmission line limit. (c) Statement and Proof of the J m Theorem The Jm Theorem. A transmission line consists of two conductors called C 2 and C3, since index 1 is reserved for the dielectric. For example, the dielectric has permeability μ1 (but we write β in place of β1). The total "particular" vector potential A z(x) due to the conduction currents in these two conductors is, according to (1.5.9), A(x)part = 1 4π Σi=23∫ Ci μiJci(x') e-jβdR R dV' . R = | x-x'| (1.5.9) The theorem claims that (1) the correct adjusted total potential is given by A(x) = 1 4π Σi=23∫ Ci [ μiJci(x') + μ0Jmi(x')] e-jβdR R dV' . (B.6.27) where Jmi(x') represents the surface current at the surface of conductor C i, and (2) this correct total potential satisfies the boundary conditions (B.6.0) at the surface of both conductors. We claim the theorem is also true for a transmission line consisting of any number of conductors, but we restrict our interest to two conductors. We shall show for (B.6.27) that (B.6.0) is valid at the surface of conductor C 2 and a similar argument then shows it is also valid at the surface of C 3. Since we are operating in the transmission line limit, the actual claim being made is this: A z(x) = 1 4π Σi=23∫ Ci [ μiJczi(x') + μ0Jmzi(x')] e-jβdR R dV' . (B.6.28) Appendix B: Magnetization Surface Currents on a Conductor 254 To enhance clarity, we shall give each conductor its own custom integration variable. The expression above contains four terms which we now write out: A z(c2)(x) = 1 4π ∫ C2 [μ2 Jcz2(x2')] e-jβdR2 R2 dV2' R2 = |x - x2'| Az(c3)(x) = 1 4π ∫ C3 [μ3 Jcz3(x3')] e-jβdR3 R3 dV3' R3 = |x - x3'| (B.6.4)' A z(m2)(x) = 1 4π ∫ C2 [μ0 Jmz2(x2')] e-jβdR2 R2 dV2' R2 = |x - x2'| Az(m3)(x) = 1 4π ∫ C3 [μ0 Jmz3(x3')] e-jβdR3 R3 dV3' R3 = |x - x3'| (B.6.5') Az(x) = Az(c2)(x) + Az(c3)(x) + Az(m2)(x) + Az(m3)(x) . (B.6.6)' Without loss of generality, we shall consider x → s where s is a point on the surface of conductor C 2. The "conduction solutions" A z(ci) (that is to say, the particular solutions) are naturally smooth at point s , as described in the text surrounding (B.6.0). Thus, we know that Az(ci)(s+) = Az(ci)(s- ) ∂nAz(ci)(s+) = ∂nAz(ci)(s-) . i = 2, 3 (B.6.7)' Since s = s+ = s- for these functions, we may trivially write 1 μ1 ∂n[Az(c2)(s+) + Az(c3)(s+)] – 1 μ2 ∂n[Az(c2)(s-) + Az(c3)(s-)] = + [ 1 μ1 - 1 μ2 ] ∂n[Az(c2)(s) + Az(c3)(s) ] . ( B . 6 . 8 ) ' Since this is non-zero, the term [A z(c2)(s+) + Az(c3)(s+)] on its own does not meet the required slope boundary condition (B.6.0) at an interface between μ1 and μ2, and that is precisely why we need the Az(m) terms. Our goal is to show that 1 μ1 ∂n[Az(m2)(s+) + Az(m3)(s+)] – 1 μ2 ∂n[Az(m2)(s-) + Az(m3)(s-)] = - [ 1 μ 1 - 1 μ2 ] ∂n[Az(c2)(s) + Az(c3)(s)] . (B.6.9)' Appendix B: Magnetization Surface Currents on a Conductor 255 so that when we add the four terms of (B.6.6)' we will get 1 μ 1 ∂nAz(s+) – 1 μ2 ∂nAz(s- ) = 0 (B.6.10) as required by (1.1.46). The concludes our proof outline , and it remains then to demonstrate (B.6.9)'. At this point, we skip over several equations of the Lemma proof since they are all generalized simply by adding 2 or 3 subscripts in the right places. For example, the surface currents are given by, K z2 = - ( μ2 μ0 - μ1 μ0 ) Hθ2 on C 2 Kz3 = - ( μ3 μ0 - μ1 μ0 ) Hθ3 on C 3 . (B.1.10) We arrive then at (B.6.16)' as follows ( recall that E 2 is the 2D Helmholtz propagator ), Az(m2)(x) = ∫{C2 ds2' [(μ1-μ2) Hθ2(x2')] E2(x|x2') Az(m3)(x) = ∫{C3 ds3' [(μ1-μ3) Hθ3(x3')] E2(x|x3') . (B.6.16)' We regard these as two representations of Stakgold's first equation (B.6.2). Nothing special happens in these equations as x → s. Thus, we can say that in either case, A z(mi)(s+) = A z(mi)(s-). When this is combined with the first line of (B.6.7)', we find by adding all four terms in (B.6.6)' that A z(s+) = Az(s-) and we have thus shown that the first boundary condition of the pair (B.6.0) is satisfied. A major difference appears at the next step (B.6.18)' , ∂nAz(m2)(s±) = ∫{C2 ds2' [(μ1-μ2) Hθ2(x')] ∂nE2(s|x2') ∓ [(μ1-μ2) Hθ2(s)]/2 . ∂ nAz(m3)(s±) = ∫{C3 ds3' [(μ1-μ3) Hθ3(x')] ∂nE2(s|x3') . (B.6.18)' The "extra Stakgold term" only appears when a point s lies on the surface being integrated over since it is this integration which gives rise to the singular situation. Since our s lies on C 2 and not on C 3, there is no "extra term" in the last equation above. Now since we want to prove (B.6.9)', we first evalua te its left hand side using each equation of (B.6.18)' twice, Appendix B: Magnetization Surface Currents on a Conductor 256 1 μ1 ∂n[Az(m2)(s+) + Az(m3)(s+)] – 1 μ2 ∂n[Az(m2)(s-) + Az(m3)(s-)] = 1 μ1 { ∫{C2 ds2' [(μ1-μ2) Hθ2(x2')] ∂nE2(s|x2') - [(μ1-μ2) Hθ2(s)]/2 } // 1 μ1 ∂n Az(m2)(s+) – 1 μ2 { ∫{C2 ds2' [(μ1-μ2) Hθ2(x2')] ∂nE2(s|x2') + [(μ1-μ2) Hθ2(s)]/2 } // - 1 μ2 ∂n Az(m2)(s-) + 1 μ1 { ∫{C3 ds3' [(μ1-μ3) Hθ3(x3')] ∂nE2(s|x3') } // 1 μ1 ∂n Az(m3)(s+) – 1 μ2 { ∫{C3 ds3' [(μ1-μ3) Hθ3(x3')] ∂nE2(s|x3') } // - 1 μ2 ∂n Az(m3)(s-) = ( μ1-μ2) [ 1 μ1 - 1 μ2 ] ∫{C2 ds2' Hθ2(x2') ∂nE2(s|x2') – [ 1 μ1 + 1 μ2 ] (μ1-μ2) Hθ2(s)/2 + ( μ1-μ3) [ 1 μ1 - 1 μ2 ] ∫{C3 ds3' Hθ3(x3')] ∂nE2(s|x3') (B.6.19)' where the last line was not present in (B.6.19). Our task of showing that (B.6.9)' is true then boils down to showing that the last expression above is equal to - [ 1 μ1 - 1 μ2 ] ∂n[Az(c2)(s) + Az(c3)(s)] . That is to say, we have to show (μ 1-μ2) [ 1 μ1 - 1 μ2 ] ∫{C2 ds2' Hθ2(x2') ∂nE2(s|x2') – [ 1 μ1 + 1 μ2 ] (μ1-μ2) Hθ2(s)/2 + (μ1-μ3) [ 1 μ1 - 1 μ2 ] ∫{C3 ds3' Hθ3(x3')] ∂nE2(s|x3') = - [ 1 μ1 - 1 μ2 ] ∂n[Az(c2)(s) + Az(c3)(s)] . ? ( B . 6 . 2 0 ) ' As before, a question mark indicates an equation that we want to show is true, but have not yet done so. Cancelling ( μ1-μ2) factors gives [ 1 μ1 - 1 μ2 ] ∫{C2 ds2' Hθ2(x2') ∂nE2(s|x2') – [ 1 μ1 + 1 μ2 ] Hθ2(s)/2 - (μ1-μ3) 1 μ1μ2 ∫{C3 ds3' Hθ3(x3')] ∂nE2(s|x3') = + 1 μ1μ2 ∂n[Az(c2)(s) + Az(c3)(s)] ? o r ( B . 6 . 2 1 ) ' (μ2-μ1) ∫{C2 ds2' Hθ2(x2') ∂nE2(s|x2') – (μ2+μ1) Hθ2(s)/2 + (μ 3-μ1) ∫{C3 ds3' Hθ3(x3')] ∂nE2(s|x3') = ∂n[Az(c2)(s) + Az(c3)(s)] . ? (B.6.22)' The integrals in (B.6.22)' can be replace d using (B.6.18)' with the s+ choice, Appendix B: Magnetization Surface Currents on a Conductor 257 ∂nAz(m2)(s+) = ∫{C2 ds2' [(μ1-μ2) Hθ2(x2')] ∂nE2(s|x2') - (1/2)[( μ1-μ2) Hθ2(s) ∂nAz(m3)(s) = ∫{C3 ds3' [(μ1-μ3) Hθ3(x2')] ∂nE2(s|x3' ) (B.6.18)+ so (μ2-μ1) ∫{C2 ds2' Hθ2(x2') ∂nE2(s|x2') = – ∂nAz(m2)(s+) - (1/2)[(μ 1-μ2) Hθ2(s) (μ3-μ1) ∫{C3 ds3' Hθ3(x3')] ∂nE2(s|x3') = – ∂nAz(m3)(s) . (B.6.23)' Equation (B.6.22)' then becomes – ∂ nAz(m2)(s+) - (1/2)[( μ1-μ2) Hθ2(s) – (μ2+μ1) Hθ2(s)/2 – ∂nAz(m3)(s) = ∂n[Az(c2)(s) + Az(c3)(s) ] ? or – ∂ nAz(m2)(s+) – μ1 Hθ2(s) – ∂nAz(m3)(s) = ∂n[Az(c2)(s) + Az(c3)(s) ] ? or - μ 1Hθ2(s) = ∂n[Az(c2)(s) + Az(c3)(s) + Az(m2)(s+) + ∂nAz(m3)(s) ] ? or - μ 1Hθ2(s) = ∂nAz(s+) ? // using (B.6.6)'. (B.6.24)' But this last equation is true as shown in Fig B.6 (with C = C 2) and (B.6.26), so we can then go back and erase all the question marks and we have then verified equation (B.6.9)' and our proof is complete. By then taking s to be a point on the surface of C 3, we would find - μ1Hθ3(s) = ∂nAz(s+) for (B.6.24)' and then Fig B.6 with C = C 3 would verify this result as well. B.7 Application of the J m Lemma to a round wire with uniform J z We shall here verify the J m Lemma for a round wire which we ca n regard as the central conductor of a coaxial transmission line with distant shield retu rn. We know from Comment 1 below (B.6.1) that the potential of the shield is not going to interfere with the J m Lemma and that a verification of the boundary conditions (B.6.0) for the potential of this Lemma applies as well to the combined potential of both conductors. Assuming the transmission line limit of small β so e-jβR ≈ 1, we start then with (B.6.1), Az(x) = ∫ C dV' [μ2 Jcz(x') + μ0 Jmz(x')] 1 4πR . R = | x - x'| (B.6.1) but we go at once to the 2D solution [ ∇2 2DA(x) = - μ2Jc(x)] limit to get Appendix B: Magnetization Surface Currents on a Conductor 258 -Az(x) = ∫ C dS' [μ2 Jcz(x') + μ0 Jmz(x')] ln(R2) 4π . // 1 2π ln(1/R) = - ln(R2) 4π (B.7.1) The two currents are given by J cz(x') = Jcz = I/(π a2) / / u n i f o r m Jmz(x') = Kz δ(r'-a) with K z = - ( μ2 μ0 - μ1 μ0 ) Hθ . (B.1.10) (B. 7.2) (a) The A z(c) term The first term in (B.7.1) is then -A z(c)(r,θ) = μ2I πa2 ∫ C dS' ln(R2) 4π = μ2I πa2 ∫0 a r' dr' ∫-π π dθ' ln(R2) 4π = μ2I 4π2a2 ∫0 a r' dr' ∫-π π dθ' ln(r'2 +r2-2rr' cos(θ-θ') ) = μ2I 2π2a2 ∫0 a r' dr' ∫0 π dx ln(r'2 +r2 - 2rr' cosx ) = μ2I 2π2a2 ∫0 a r' dr' Q(r',r) (B.7.3) where we have defined the integral Q(r',r) ≡ ∫0 π dx ln [r'2 +r2-2rr' cosx] . (B.7.4) The round wire geometry is shown in this drawing, where R2 shown above comes from the law of cosines, Fig B.7 Appendix B: Magnetization Surface Currents on a Conductor 259 The integral Q(r',r) may be evaluated using GR7 p 531 4.224, with a = r'2 +r2 and b = -2rr' and a2-b2 = (r'2-r2)2 so that a2-b2 = | r'2-r2 | . The condition a > |b| > 0 is met since (r±r')2 > 0 => r2+r'2 > ±2rr' which says a > ±b so a > |b|. Thus, Q(r',r) = ∫0 π dx ln [r'2 +r2-2rr' cosx)] = π ln [ (r'2 + r2 )+ | r'2- r2 | 2 ] = ⎩⎨⎧ π ln [r'2] r' > r π ln [r2] r' < r = 2 π ⎩⎨⎧ ln(r') r' > r ln(r) r' < r . ( B . 7 . 5 ) We then have -Az(c)(r,θ) = μ2I 2π2a2 ∫0 a dr' r' Q(r',r) = μ2I πa2 ∫0 a dr'r' ⎩⎨⎧ ln(r') r' > r ln(r) r' < r = μ2I πa2 ∫0 a dr' r' { lnr' θ(r'>r) + lnr θ(r'<r) } = μ2I πa2 [ θ(r<a) ∫r a dr' r' lnr' + lnr ∫0 min(a,r) dr' r' ] = μ2I πa2 { θ(r<a) (1/2){a2lna - r2lnr - (a2-r2)/2} + (1/2)lnr [min(a,r)]2 } (B.7.6) where Maple says . We then write out A z(c)(r) in its two regions Az(c)(r>a) = - μ2I πa2 { (1/2) a2lnr } = - μ2I 2π lnr (B.7.7) Az(c)(r<a) = - μ2I πa2 { (1/2){a2lna - r2lnr - (a2-r2)/2} + (1/2) r2 lnr} = μ2I 2πa2 { (a2-r2)/2 - a2lna } Appendix B: Magnetization Surface Currents on a Conductor 260 (b) The A z(m) term From (B.7.1) and (B.7.2), -A z(m)(x) = μ0∫ C dS' Jmz(x')] ln(R2) 4π = μ0 4π ∫0 a r' dr' ∫-π π dθ' [- ( μ2 μ0 - μ1 μ0 ) Hθ(r') ] δ (r'-a) ln(R2) = - a 4π Hθ(a) (μ2- μ1) ∫-π π dθ' ln(a2 +r2-2ra cos(θ-θ')) = - a 2π Hθ(a) (μ2- μ1) ∫0 π dx ln(a2 +r2-2ra cos(x)) = - a 2π Hθ(a) (μ2- μ1) Q(a,r) = - a 2π Hθ(a) (μ2- μ1) 2π ⎩⎨⎧ ln(a) a > r ln(r) a < r // using (B.7.5) = a H θ(a) (μ1- μ2) ⎩⎨⎧ ln(a) a > r ln(r) a < r . (B.7.8) From Ampere's law (1.1.37) we have (ignoring displacement current inside the conductor) ∫{ H • ds = 2πaHθ(a) = ∫S J • dS = I => H θ(a) = I 2πa and so -A z(m)(x) = I 2π (μ1-μ2) ⎩⎨⎧ ln(a) a > r ln(r) a < r . ( B . 7 . 9 ) Then Az(m)(r>a) = I 2π (μ2-μ1) lnr Az(m)(r<a) = I 2π (μ2-μ1) l n a . ( B . 7 . 1 0 ) (c) Adding the two terms and checking boundary conditions Adding t he results of (B.7.7) and (B.7.10) we obtain the total A z vector potential, Az(r>a) = - μ2I 2π lnr + I 2π (μ2- μ1) lnr = - μ1I 2π lnr Az(r<a) = μ2I 2πa2 { (a2-r2)/2 - a2lna } + I 2π (μ2-μ1) lna Appendix B: Magnetization Surface Currents on a Conductor 261 = μ2I 2π { (a2-r2)/(2a2) - lna } + I 2π (μ2-μ1) lna = - I 2π { μ2 (r2-a2)/(2a2) + μ1 lna } Here then are the finally results for the potential A z = Az(c) + Az(m), Az(r>a) = - I 2π μ1lnr Az(r<a) = - I 2π { μ1 lna + μ 2r2-a2 2a2 } . ( B . 7 . 1 1 ) As a check, we calculate the B and H fields implied by these potentials B = curl A = r^ [ r-1∂θAz - ∂zAθ] + θ^ [∂zAr - ∂rAz] + z^ [ r-1∂r(rAθ) - r-1∂θAr ] = θ^ [- ∂ rAz] => B θ = -∂rAz(r) Then Bθ(r>a) = I 2π μ1 (1/r) Bθ(r<a) = I 2π μ2 2r 2a2 = I 2π μ2 (r/a2) ( B . 7 . 1 2 ) so the H fields are then Hθ(r>a) = I 2π (1/r) Hθ(r<a) = I 2π (r/a2) . ( B . 7 . 1 3 ) This agrees with Ampere's Law applied in these two regions: 2πr H θ(r>a) = I => H θ(r>a) = I 2π (1/r) 2πr Hθ(r<a) = I( πr2/πa2) => H θ(r>a) = I 2π (r/a2) . (B.7.14) Next, we check the two boundary conditions required by (B.6.0) : value at r = a: Az(r>a)|r=a – Az(r<a) |r=a = - I 2π μ1lna - (I 2π [ μ2 a2-a2 2a2 - μ1 lna ]) = 0 OK Appendix B: Magnetization Surface Currents on a Conductor 262 slope at r = a: [ ∂rAz = -Bθ so use (B.7.12) ] ∂rAz(r>a) |r=a = - I 2π μ1 (1/a) ∂rAz(r<a) |r=a = - I 2π μ2(a/a2) = - I 2π μ2(1/a) 1 μ1 ∂rAz(r>a) |r=a – 1 μ2 ∂rAz(r<a) |r=a = - I 2π (1/a) - [- I 2π (1/a)] = 0 OK Thus we have shown for the round conductor with uniform J z that "the J m Lemma" works. By adding the bogus surface current term, we generate the correct homogeneous solution which when added to the Helmholtz integral provides the correct total solution which meets both boundary conditions. (d) Plots of A z and B θ and H θ Maple provides plots of A z from (B.7.11), B θ from (B.7.12) and H θ from (B.7.13) for this round wire situation. Parameters are set to I = 1, a = 2, μ1 = 2 (dielectric), μ2 = 3 (wire). Appendix B: Magnetization Surface Currents on a Conductor 263 Fig B.8 Az wanders down as ~ -ln(r) for large r; B θ jumps down at r = a while H θ is continuous there. Appendix C: DC Properties of a Wire 264 Appendix C: DC Properties of a Wire C.1 The DC resistance of a wire The resistanc e per unit length R of a differential piece of wire of length dz and area dA is derived as follows ( σ = conductivity), dV = E dz, J = σ E, I = J dA ⇒ Rdz = dV/I = Edz/JdA = (1/ σ) dz/dA ⇒ R = (1/σ) /dA . If current density J is constant across the wire cross se ction (which is the case at DC), we repeat the above with dA → A where A is the total wire cross sectional area to find this resistance per unit length for the wire, R = 1/( σA) = ρ/A // ρ = 1/σ = resistivity (C.1.1) For a round wire of radius a, A = π a 2, so R = 1/( σπa2) ≡ Rdc. Current density J is uniform at DC because there are no eddy currents to make it non-uniform as d escribed in Appendix P. We of course assume that the wire is made of an isotropi c and homogeneous substance where σ is the same in all directions at at all points inside the wire. C.2 The DC surface impedance of a wire I magine a fat wire carrying current I, If, at the surface of the wire, one puts voltmeter probes at longitudinal spacing dz, one measures some potential difference which is dV = E zdz. When probed at the surface, the wire appears to have this impedance, (Zsdz) = dV/I . The quantity Z s is the surface impedance per unit length and is thus given by Z s = Ez / I ( C . 2 . 1 ) where E z is the component of electric field at the surface in the direction of the wire. For a wire operating at DC, the current density is uniform across the wire so J z = I/A and E z = Jz/σ = I/(Aσ). Thus, Z s = 1/(Aσ) = ρ /A // = R of (C.1.1) (C.2.2) Appendix C: DC Properties of a Wire 265 where A is the cross sectional area. For a round wire of radius a, A = πa2, so Z s = ρ/(πa2) . ( C . 2 . 3 ) If the wire is a perfect conductor, ρ = 0 and Z s = 0. For DC, we have Z s = R, but for AC this is no longer true due to the skin and proximity effects which make J z non-uniform over the conductor cross section. Again, see Appendix P for a general discussion of both these effects, and Chapter 2 for skin effect. C.3 The DC internal and external inductance of a round wire We assu me here that μi is the internal permeability of the wire, and μe of the region external to the wire. The energy density (joules/m 3) stored in an electromagnetic field w ithin a medium of negligible loss is given by u em = (E•D + B•H)/2 [ Jackson p 259 Eq. (6.106) ] . We are interested only in the portion of this energy density stored in the magnetic field, so u = (1/2) B•H. Since μ and ε are assumed to be scalars, u = (1/2) μ H 2. ( C . 3 . 1 ) For any inductor of inductance L carrying current I and having potential difference V, we know that V = L d I / d t ( C . 3 . 2 ) P = IV = I (L dI/dt) = d/dt [ (1/2)L I 2 ] . ( C . 3 . 3 ) Since the power fed into an ideal inductor goes into the magnetic field energy, P = dU/dt and so U = (1/2)L I2 . ( C . 3 . 4 ) For a straight wire, we now redefine symbol L to mean inductance per unit length, so then U = (1/2)[Ldz] I2 . ( C . 3 . 5 ) For either the internal or external region we have U = ∫dV u = dz ∫dS u where d V is a volume element and dS is a cross sectional area element. Thus from (C.3.1) and (C.3.5), U = dz∫dS (1/2) μ H2 = (1/2)[Ldz] I2 => L = μ ∫dS (H/I)2 . (C.3.6) In particular [ elsewhere we have used μi = μ and μe = μd ] Li = μi ∫in dS (Hi/I)2 ( C . 3 . 7 ) Le = μe ∫outdS (He/I)2 . ( C . 3 . 8 ) Appendix C: DC Properties of a Wire 266 For the round wire at DC it is a simple matter to compute H i and He using Ampere's Law (1.1.37), ∫{ H • ds = ∫S J • dS 2πr Hi = I(πr2/πa2) => H i/I = 1 2π (r/a2) 2πr He = I => H e/I = 1 2π (1/r) . (C.3.9) We then compute the two inductances as follows: L i = μi ∫dS (Hi/I)2 = μi ∫0 a rdr ∫-π π dθ [1 2π (r/a2)]2 = μi 2πa4 ∫0 a r3dr = μi 8π Le = μi ∫dS (He/I)2 = μe ∫a ∞ rdr ∫-π π dθ [1 2π (1/r)]2 = μe 2π ∫a ∞ (1/r) dr = ∞ . Both results are interesting. L i is interesting because it is independent of the wire radius a. For the same current, a smaller a results in a larger H and B field, which is then offset by the smaller volume (area). L e is interesting because it is infinite ! Even a tiny 1 cm piece of our infinitely long round wire stores an infinite amount of energy in its magnetic field. We therefore limit the external region by some large radius R and then we have Li = μi 8π = μi μ0 μ0 8π = μi μ0 4π x 10-7 8π = μi μ0 * 50 nH/m (C.3.10) Le = μe 2π l n ( R / a ) . ( C . 3 . 1 1 ) Thus for a non-magnetic round wire in air the internal inductance is exactly 50 nH/m. A "practical wire" is more like a loop of wire than an infinitely long wire. It is difficult to conjure up an experiment to test (C.3.11) even for a very long straight piece of wire without having some return path for the current to return to the driving "battery". For wire and dielectric both having μ 0, Jackson shows (p 216-218) that the inductance per unit length of a loop of projected area A of radius-a wire is given by Le + Li = (μ0/4π) [ ln(ξ A/a2) + 1/2], where ξ is a near-unity factor which accounts for messy details of the calculation. The 1/2 term accounts for the internal inductance L i = μ0/8π as in (C.3.10). For a circular loop of radius R, one has A = π R 2 and, if R >> a, ξ = 64/(πe4) ≈ .373. So, ln(ξA/a2) = ln(64 πR2/πe4a2) = 2 ln(8R/ae2) = 2 ln(8R/a) + 2 ln(e-2) = 2 ln(8R/a) - 4 and then the total inductance per unit length is L e + Li = (μ0/4π) [2 ln(8R/a) - 4 + 1/2] = ( μ0/2π) [ ln(8R/a) - 2 + 1/4] = ( μ0/2π) [ ln(8R/a) -7/4] . Appendix C: DC Properties of a Wire 267 The total inductance of such a loop is then L = 2 πR(μ0/2π) [ ln(8R/a) -7/4] = μ 0R [ ln(8R/a) -7/4] (C.3.12) in agreement with Jackson Problem 5.32 p 234. If one omits the internal inductance, the last factor is -2 instead of -7/4, and this result is seen in some sources. The point is that this is a finite result, even though the (dipole) magnetic field of such a loop extends to infinity. A loop of N turns gets an extra factor N2 because in effect current I → NI in (C.3.5), so the total field energy increases by factor N2. Suppose there were two parallel wires with currents flow ing in opposite directions. In this case, we could compute the magnetic field H at any point in space as the vector sum of the fields of the two wires, then we could integrate H2 over all space to get the total energy U and from that the external inductance L e. In this case, the ln(R) divergence does not appear. In ef fect, the divergence cancels between the two wires, similar to the way opposite short segments of the circular wire cancel to give the finite result quoted above. We really only care about the internal inductance L i of a wire in our transmission line analysis because the external inductance L e is already accounted for by the techniques of Chapter 4. That is, L e is computed by considering the magnetic potential A z ( or W ) between the wires, see (4.8.10). Appendix C: DC Properties of a Wire 268 C.4 The DC internal inductance of a wire of rectangular cross section The inductance expression s above apply to any cross sectional shape, Li = μi ∫in dS (Hi/I)2 (C.3.7) Le = μe ∫outdS (He/I)2 (C.3.8) so the only problem is how to compute H for a non -round wire. That problem is solved in Appendix B where it is shown that H( x,y) = - 1 4π ∫d2x' ln(R2) curl' J (x') R = | x-x'| . (B.2.6) (C.4.1) Consider a wire of rectangular cross section 2a x 2b (uniform J z) as an example. Fig C.1 The current density is given by J z(x) = (I/4ab) θ(-a ≤ x ≤ a) θ(-b ≤ y ≤ b) = (I/4ab) θ(x ≤ a)θ(x≥-a) θ(y ≤b)θ(y≥-b) // θ(s≥r) means θ(s-r) Heaviside = (I/4ab) θ(a- x)θ(x + a) θ (b- y)θ (y +b) . (C.4.2) Calculate curl J : curl J = x^ (∂ yJz - ∂zJy) + y^ (∂zJx - ∂xJz) + z^ (∂xJy - ∂yJx) = x^ (∂ yJz) + y^ (- ∂xJz) ( C . 4 . 3 ) ∂xJz = (I/4ab) ∂x[θ(a- x)θ(x + a)] θ (b- y)θ(y +b) = (I/4ab) [ θ(a-x)δ (x+a) - δ (x-a) θ(x+a)] θ(b- y)θ (y +b) // ∂xθ(a-x) = - δ(x-a) ∂ yJz = (I/4ab) θ(a- x)θ(x + a) ∂y[θ(b- y)θ(y +b)] = (I/4ab) θ(a- x)θ(x + a) [θ(b- y)δ (y+b) - δ(y-b) θ(y +b)] Appendix C: DC Properties of a Wire 269 so then [curl J]x = (I/4ab) θ(a- x)θ(x + a) [θ(b- y)δ (y+b) - δ (y-b) θ(y +b)] [curl J]y = - (I/4ab) [ θ(a-x)δ(x+a) - δ (x-a) θ(x+a)] θ(b- y)θ (y +b) . (C.4.4) Notice that we can obtain [curl J]y from [curl J]x by doing a ↔b, x↔y and adding a minus sign. Finally calculate H x from (C.4.1). Hx(x,y) = - 1 4π ∫d2x' ln(R2) [curl' J (x')]x R = | x-x'| = - I 16πab ∫-a a dx' ∫-∞ ∞ dy' ln [ (x-x')2 + (y-y')2] [θ(b - y')δ (y'+b) - θ(y'+b)δ(y'- b) ] } = - I 16πab ∫-a a dx' ∫-∞ b dy' ln [ (x-x')2 + (y-y')2] δ(y' +b) + I 16πab ∫-a a dx' ∫-b ∞ dy' ln [ (x-x')2 + (y-y')2] δ(y'- b) = - I 16πab ∫-a a dx' ln [ (x'-x)2 + (y+b)2] + 1 16πab ∫-a a dx' ln [ (x'-x)2 + (y-b)2] } ≡ I 16πab ( -I1+I2) I1(b) = ∫-a a dx' ln [ (x'-x)2 + (y+b)2] I 2(b) ≡ ∫-a a dx' ln [ (x'-x)2 + (y-b)2] = I1(-b) . (C.4.5) ≡ I 16πab F(x,y,a,b) F = -I 1+ I2 . As an aid to Maple's grouping of elements, let x" = x'-x, then take x" → x' to get I1(b) = ∫-a-x a-x dx' ln [ x'2 + c2] where c = y+b . Maple then evaluates I 1 and I2 as follows. Appendix C: DC Properties of a Wire 270 In Maple, unapply(f,x) causes expression f to be a func tion of x which can then be called as f(x). Collect just orders terms in a certain way, while subs fo rces Maple to be a little smarter about expressions. The next step is to create function F(x,y,a,b) which is just -I 1+ I2 as shown above Reading from the above and putting x,y last in each parentheses, the four arctangents can be written (1/2) F(x,y,a,b) atan = -(b-y) [- tan-1 a+x b-y ] +(b-y)tan-1a-x b-y - (b+y)tan-1(a+x b+y ) +(y+b)[-tan-1(a-x b+y )] = (b-y) [ tan-1 a+x b-y + tan-1a-x b-y ] - (b+y) [ tan-1(a+x b+y ) + tan-1(a-x b+y ) ] . Next, the four log terms can be combined to give (1/2) F(x,y,a,b) ln = (1/2) (a-x) ln [ (a-x)2 +(b-y)2 (a-x)2 +(b+y)2 ] + (1/2) (a+x) ln [ (a+x)2 +(b-y)2 (a+x)2 +(b+y)2 ] . Combining and reordering these terms, we get Appendix C: DC Properties of a Wire 271 (1/2) F(x,y,a,b) = (1/2) (a+x) ln [ (a+x)2 +(b-y)2 (a+x)2 +(b+y)2 ] + (1/2) (a-x) ln [ (a-x)2 +(b-y)2 (a-x)2 +(b+y)2 ] + (b-y) [ tan-1a-x b-y + tan-1 a+x b-y ] - (b+y) [ tan-1(a-x b+y ) + tan-1(a+x b+y ) ] . (C.4.6) This expression agrees with Holloway and Kuester's W 1 if one replaces a = w/2 and b = t/2. With such replacements, we would have Hx(x,y) = I 16πab F(x,y,a,b) = I 4πwt F(x,y, w/2, t/2) = I 2πwt [ (1/2) F(x,y, w/2, t/2) ] ≡ I 2πwt W1 ( C . 4 . 7 a ) so our H x then agrees with their equations (9) and (11). Now based on the comment below (C.4 .4) above, we may conclude for H y that Hy(x,y) = - 1 4π ∫d2x' ln(R2) [curl' J (x')]y = Hx(x,y) if we swap a ↔b, x↔y and add a minus = - 1 16πab F ( y , x , b , a ) . ( C . 4 . 7 b ) Just for the record, We can now make a "field plot" showing the H fiel d (direction and magnitude) in the cross section plane of our rectangular conductor, where we stick with the 4:1 ratio of edges as in Fig C.1 above, Appendix C: DC Properties of a Wire 272 F i g C . 2 In this plot the H field appears to be maximal at the conductor boundary (shown in red) and as one moves away it becomes the field of a thin round wire. Current J z is flowing in the z direction toward the viewer. The second plot is of | H|2 as a surface over the x,y plane. Recall that | H|2 is proportional to the energy density in the magnetic field which in turn contributes to inductance. Appendix C: DC Properties of a Wire 273 view from above view from below F i g C . 3 The | H| 2 surface is very steep at the conductor boundaries, somewhat resembling a rectangular volcano which dips all the way down to 0 in the center, as shown on the right. The L i integration discussed below is over this central "cone" of the volcano. The red plot below is a slice through the volcano at x = 0 : Fig C.4 The black plot is of | H | (but scaled down) and resembles the H θ plot for the round wire shown in Fig B.8. We return now to a computation of the internal inductance Li of the rectangular wire. From (C.3.7), Appendix C: DC Properties of a Wire 274 Li = μi ∫idS (H/I)2 = μi (1 16πab )2 ∫-a a dx ∫-b b dy [F(x,y,a,b)2 + F(y,x,b,a)2] . (C.4.8) Since F(x,y, αa,αb) = αF(x/α,y/α,a,b) one can show that L i must have this functional form Li = (μi/8π) f( b / a ) ( C . 4 . 9 ) though this conclusion is obvious based on dimensions alone. The factor ( μi/8π) is Li for a round wire of any radius, as shown in (C.3.10) . The problem is to find function f . We set a = 1 with no loss of generality, and use the obvious four-fold symmetry of the energy density so that Li = 4 μi (1 16πb )2 ∫01 dx ∫0 b dy [F(x,y,1,b)2 + F(y,x,b,1)2] = ( μi 8π ) { 1 8π2b2 ∫01 dx ∫0 b dy [F(x,y,1,b)2 + F(y,x,b,1)2] } . (C.4.10) Thus our function of interest is f(b) = 1 8π 2b2 ∫01 dx ∫0 b dy [F(x,y,1,b)2 + F(y,x,b,1)2] (C.4.11) where If one were to expand this expression, there would be 162 + 162 = 256 + 256 = 512 terms if no terms combined. In fact there are 232 terms: Here are four sample terms in the integrand of (C.4.11), Appendix C: DC Properties of a Wire 275 It seems rather unlikely that all 232 terms can be doubl e-integrated analytically! For example, if we ask Maple to analytically integrate the last term shown above just over the x range, it gives up, Thus, in order to compute f(b) we must turn to numerical integrati on which, for each value of b, requires doing 232 numerical double integrals and adding up the results. As is visible in Fig C.3 and Fig C.4, the overall integrand is singular at the conductor edge, so we might expect some difficulties with the numeric integrations near the upper endpoints. We were not successful trying for a hour to get Maple to compute the integral (C.4.11) analytically or numerically, but certainly the numerical integration can be done. Holloway and Kuester quote the following numerical approximate formulas for two special cases, L i = (μi/8π) [0.96639] a = b (square wire) // very close to the round conductor L i = (μi/8π) [(4π/3) b/a ] b/a << 1 (flat wire) // L i = (1/6) μi (b/a) (C.4.12) We discuss the second case in Section C.5 Reader Exercise: Do the numerical integration outli ned above to determine function f(b) for several b values and plot for b = 1 to 10. Is f(1) = 0.96639 ? Holloway and Kuester have a plot in their Fig 2 which looks like this for L i [Li for a circular wire is 50 nH as shown in (C.3.10)], Fig C.5 Appendix C: DC Properties of a Wire 276 Comment: Appendix B.2 provides three methods of computing H from J . We chose to use the formula (B.2.6). Holloway and Kuester use the second method of first computing A then B = curl A. A third method is to use the 2D Biot-Savart Law (B.2 .24). That third method begins this way : H(x,y) = ∫d2x' 1 2πR2 J(x') x R R ≡ x - x' (B.2.24) and Jz(x) = (I/4ab) θ(-a ≤ x ≤ a) θ(-b ≤ y ≤ b) so H(x,y) = ∫-a a dx' ∫-b b dy' 1 2πR2 (I/4ab) z^ x R . But R = (x-x') x^ + (y-y') y^ => z ^ x R = (x-x') y^ - (y-y') x^ . Thus, Hx(x,y) = ∫-a a dx' ∫-b b dy' 1 2πR2 (I/4ab) [- (y-y') ] Hy(x,y) = ∫-a a dx' ∫-b b dy' 1 2πR2 (I/4ab) [+ (x-x') ] or Hx(x,y) = (I/8 πab) ∫-a a dx' ∫-b b dy' (y'-y)/R2 Hy(x,y) = - (I/8 πab) ∫-a a dx' ∫-b b dy' (x'-x)/R2 . (C.4.13) These last integrals are the same as (7) and (8) of Holloway and Kuester with 2a = w and 2b = t. C.5 The DC internal inductance of a thin flat wire This is a fasci nating problem with a result that is non-obvious. Consider an infinitely long conductor whose cross se ction has the shape of a thin strip of width w and height t with t << w, Fig C.6 The correct result for L i was given earlier in (C.4.12) and we repeat it here, setting w = 2a and t = 2b, L i = (μi/8π) [(4π/3) t/w ] = (1/6) μi (t/w) t << w . (C.4.12) We shall now attempt to obtain this result in a simple manner, intentionally misleading the reader a bit. Appendix C: DC Properties of a Wire 277 The uniform current density is J z = I/(wt), flowing toward the viewer. Here is a blowup of a piece of the strip near its center, Fig C.7 The red math loop is positioned as shown fo r an application of Ampere's Law, ∫S J • dS = ∫{C H • ds . (1.1.37) Starting at the lower left corner of the red loop for ∫{C, this says Jz 2y s = s H x(-y) + 0 2y - H x(y)s - 0 2y . (C.5.1) We assume that "near the center of the strip" there is no significant transverse field component H y, though we accept that such transverse fields do exist far aw ay "near the edges" of the strip as in Fig C.2, Fig C.8 Thus, the two vertical sections of th e red loop make negligible contribution to the line integral in the main central region. Symmetry indicates that H x on the upper red loop segment is equal and opposite to that on the lower segment, so the lin e integral is then -2H x(y) s and we continue : Jz 2y s = - 2 H x(y)s J z y = - Hx(y)s Appendix C: DC Properties of a Wire 278 I/(wt)*y = - H x(y)s Hx(y)/I = - y/(wt) . (C.5.2) The result is that H x(y) = -(I/wt)y which is a very reasonable linear function of y, with H x(y=0) = 0. The fact that H x(y) does not depend on x is also reasonable since, when w >> t, the central region of the strip is basically all of the strip excluding the tiny end regions which we ignore. A similar argument is made for the analysis of a parallel plate capacitor, where the end effects are ignored if w >> t. To get the internal inductance due to this H x energy storage, we compute its contribution from the dotted rectangle in Fig C.6, then multiply by (w/s) to get L i for the entire strip. So, using (C.3.7), Li = (w/s) μ ∫dotted dS (H/I)2 = (w/s) μ∫dotted (sdy) [-y/(wt)]2 = (w/s) μ s (wt)-2 ∫-t/2 t/2 dy y2 = (w/s) μ s (wt)-2 2 ∫0 t/2 dy y2 = (w/s) μ s (wd)-2 2 (1/3) (t/2)3 = w-1 μ t-2 (2/3) t3/8 = (1/12) μ (t/w) = ( μ i/8π) [ 2π/3 (t/w) ] // strip w>>t , due to H x (C.5.3) But this is only half the correct result for L i which was just quoted above. By luck, (C.5.3) happens to be the correct result for the H x contribution to L i; by luck because it is derived from Fig C.7 with the assumption that H y ≡ 0 which is not true. The other half of L i in fact comes from the H y field in the strip. One can write Li = μ ∫-t/2 t/2 dy ∫-w/2 w/2 dx [ (Hx/I)2 + (Hy/I)2 ] = Lix + Liy (C.5.4) so the H x and Hy contributions are simply additive with no interference. So where did the argument above go wrong? It all s eemed so reasonable. One is of course biased by the appearance of the fields in Fig C.8 shown just a bove. One's impression is that as the aspect ratio is increased from 4:1 to perhaps 100:1, the nature of the above plot should become even more convincing: large horizontal arrows to the left along the top of th e strip, large horizontal arrows to the right along the bottom of the strip, and some minor edge effects at the distant ends. But this is in fact not a correct impression! For a 10:1 aspect ratio strip, here is the field map (H x,Hy) for the upper right quadrant of the strip Fig C.9 Appendix C: DC Properties of a Wire 279 If we plot only the H x component by setting H y = 0 in our field plot, we get Fig C.10 and this displays our conjectured functional shape H x(y) = -(I/wt)y applying not just at the center of the strip, but all along the strip. This then explains graphically why our calculation above came up with the correct result for the H x contribution to L i. The other half of L i comes from the transverse field component H y which has this appearance (we now set H x = 0 in the field plot) Fig C.11 The fact that the H y contribution to L i is exactly equal to the H x contribution is just not obvious. One would think some simple argument could be concoc ted to explain this fact. For example, one might conjecture looking at the above plot that H y ≈ Hy(x) so that Ampere's law for the Fig C.7 red loop says, using s = dx, J z 2y dx = dx H x(-y) + H y(x+dx) 2y - H x(y)dx - H y(x)2y . J z 2y = -2 H x(y) + [H y(x+dx) - H y(x)]/dx * 2y . J z y = - H x(y) + y ∂ xHy( x ) . ( C . 5 . 5 ) We might then try H x(y) = -A(I/wt)y based on Fig C.10 with A some constant. Then (C.5.5) says (I/wt) (1-A) = ∂ xHy(x) => H y(x) = [(I/ωt) (1-A)] x ≡ Bx (C.5.6) which seems reasonable in terms Figure C.11. But H y(x) = Bx is problematical in two respects: (1) there is no obvious way to determine B without taking a limit of the complicated full H y formula ; (2) Even when that is done, H y(x) is in fact not linear in x as a simple plot shows, so the model is inaccurate and does not give the result that L i due to H y is (1/12) μ (t/w). The field plots shown above and the | H| 2 energy plots of Fig C.3 are easy to produce from Maple. These latter plots are like topographical maps and they can be displayed in that manner as shown on the right below Appendix C: DC Properties of a Wire 280 Fig C.12 where we have reverted to the 4:1 aspect ratio strip. One should not confuse the H| 2 topo contour lines shown here with a plot of the H field lines. Making a field line plot is not a built-in function for our old Maple V and requires some minor coding to implement. Here is such a field line plot for a 20:1 aspect ratio thin strip (method given in Appendix O), F i g C . 1 3 Each contour starts at x = 0 and y = some value and is iterated CCW (chasing the direction of the H vector) until it arrives back where it started. The little jogs at the top represent the small error of this numerical process. Looking at this field line plot, it is totally obvious that there does not exist some "broad central region" in the st rip where the field lines are mostly horizontal. The transverse field components (vertical) appear as soon as one leaves the ex act center of the strip and it is totally wrong to ignore such transverse H y components in the computation of L i. The correct calculation of L i is done in the 2009 paper of Holloway and Kuester. They point out errors made by earlier authors and make the point that half the internal inducta nce comes from each field component. They do not claim that the transverse contribution is exactly half the result, but that it is half to a high degree of precision. In an email communication, Prof. Kuester made the appropriate point that, since div H = 0 (there is no magnetic charge), the H fiel d lines must close on themselves and that is what forces the above figure to have the shape it has, where there is no "broad central region" having essentially horizontal field lines. In the corresponding parallel plate capacitor picture for electrostatics, Appendix C: DC Properties of a Wire 281 since electric charge does exist, the E fields lines do not need to close on themselves, and have sources and sinks all along the capacitor cross section, allowing for a uniform broad central region. Here is one more field line plot showing a larger range of field lines, F i g C . 1 4 As the field lines are continued outward, they eventua lly become circles as the strip eventually becomes a line source ( a point source in cross section) when viewed from far away. Reader Exercise: Come up with a simple explanation for why the H x and Hy fields each contribute half the total L i value for the strip. Is this perhaps true for any edge ratio of the rectangular cross section? That certainly seems unlikely. C.6 The DC internal inductance of a hollow round wire The pipe geometry is as follows, where we assume J z is uniform and total current is I : Fig C.15 Appendix C: DC Properties of a Wire 282 As with the round wire case, we can avoid using (C.4.1) (or alternates) to compute H due to symmetry. The total current enclosed with in the red circle is this, Ienc(r) = area inner annulus a to r area full annulus a to b = r2- a2 b2-a2 I valid for a ≤ r ≤ b . (C.6.1) For r < a, I enc(r) = 0, and for r > b, I enc(r) = I. Ampere's Law says 2πr H θ(r) = Ienc(r) . (C.6.2) Note that H θ(r) = 0 inside the tube, so this region makes no contribution to L e or Li. Inside the annulus, Hθ(r) = Ienc(r)/ (2πr) = 1 2π 1 r r2- a2 b2-a2 I . a ≤ r ≤b (C.6.3) Recalling that Li = μi ∫in dS (H/I)2 (C.3.7) we conclude that L i = μi 1 (2π)2 1 (b2-a2)2 ∫in dS 1 r2 (r2 - a2)2 and then using dS = 2π rdr we find Li = μi 1 2π 1 (b2-a2)2 ∫a b dr 1 r (r2 - a2)2 = μi 1 2π 1 (b2-a2)2 J . (C.6.4) Maple computes the integral J as follows Appendix C: DC Properties of a Wire 283 which we restate as J = (1/4)b4 + (3/4)a4 - a2b2 + a4 ln(b/a) (C.6.5) and then the final result for the internal inductance of a hollow pipe with b > a is Li = μi 1 2π 1 (b2-a2)2 [(1/4)b4 + (3/4)a4 - a2b2 + a4 ln(b/a)] . (C.6.6) (a) limit as a → 0: should be round wire of radius b Reading off this limit from (C.6.6), Li = μi 1 2π 1 (b2-a2)2 [(1/4)b4 + (3/4)a4 - a2b2 + a4 ln(b/a)] = μi 1 2π 1 b4 [(1/4)b4 + 0 - 0 + 0 ln(b/a)] = μi 1 8π ( C . 6 . 7 ) and we recover the L i of a round wire as found in (C.3.10). (b) External Verification of (C.6.6) The result appears in a very fat (2,263 pages) 1922 handbook edited by Pender and Del Mar, from which we quote via Google books, page 827, Appendix C: DC Properties of a Wire 284 In their version of cgs units, the round wire has L i = (μ/2) per unit length according to (13a), so one must add (1/4π ) to their (14) result to compare with (C.6.6). Using r 2 = b and r 1 = a the results then agree after some algebra. (c) limit as b-a → 0: thin shell radius a and thickness d In this limit, the hollow pipe is a thin cylindrical shell of inner radius a and thickness d. Continuing the Maple code, we first replace parameter b by a+d, Maple then expands this L i function about d = 0, Of course our only interest is in the first term, so in this limit we have found that L i = μi 1 6π (d/a) = μi 8π [ (4/3)(d/a) ] thin shell, valid for d << a (C.6.8) where again ( μi/8π) is Li for a round conductor of any radius. Appendix C: DC Properties of a Wire 285 (d) Maple Plot Write (C.6.6) as L i = μi 8π { 1 (b2-a2)2 [b4 + 3a4 - 4a2b2 + 4a4 ln(b/a)] } (C.6.6) = μi 8π { 1 (1-x2)2 [ 1 + 3x4 - 4x2 - 4x4 ln(x) ] x ≡ a/b = μi 8π f(x) . Maple then plots f(x) : Fig C.16 As the pipe is hollowed out from a round wire to a foil shell, f(x) drops from 1 to 0 as shown. Appendix D: Fields inside a Round Wire 286 Appendix D: The General E and B Fields Inside an Infinite Straight Round Wire This Appendix presents a rather lengthy calculation of the fields and currents inside a round wire without the Chapter 2 assumption that such fields and curren ts are symmetrical about the axis. This wire is regarded as one conductor of an infinite transmission line down which a wave is propagating. Since this Appendix is quite long, a brief summary is in order (see also Table of Contents) : Section D.1 (a) A longitudinal traveling wave form E(r, θ,z,t) = ej(ωt-kz) E(r,θ) is assumed inside the round wire and E(r,θ,z,t) is then shown to satisfy a certain vector Helmholtz equation. (b) The field E(r,θ) and the surface charge n( θ) are both expanded onto " azimuthal partial waves" ejmθ with coefficients E(r,m) and N m. (c) The Helmholtz equation's vector Laplacian @ ≡ ∇2 is stated in cylindrical coordinates. (d) The three Helmholtz component equations and div E = 0 are written out in these coordinates. Section D.2 (a),(b),(c): The z and r Helmholtz equations and the div E = 0 equation are solved for E z, then Er, and then E θ. These solutions are expressed in terms of Bessel J functions of a complex argument and two unknown constants a m and Km for each partial wave. (d) a boundary condition relating E r to surface charge density n( θ) is derived (see D.9 below) (e) this and anothe r boundary condition E θ(a,m) = 0 (see D.8 below) are used to evaluate a m and Km and then the solution E field components are stated in box (D.2.33). Section D.3 It is noted that the boxed E field solutions also satisfy the ignored third θ Helmholtz equation. Section D.4 The B fields are computed from the E fields using Maxwell -j ωB = curl E, and then box (D.4.9) summarizes both the E and B partial wave fields inside a round wire. Section D.5 These E and B fields are shown to exactly solve the other three Maxwell equations. Section D.6 The m = 0 partial wave results are stated and compared to the results of Chapter 2. Section D.7 The problem of finding an exterior field solution for the round wire is discussed. Section D.8 Arguments supporting the second boundary condition E θ(a,m) = 0 are presented. Section D.9 The "charge pumping boundary condition" is discussed in relation to surface currents. Section D.10 High frequency limits of the round wire E fields are presented. Section D.11 Low frequency limits of the round wire E fi elds are presented, along with comments on accuracy, the meaning of symbol k, and the e-jkz ansatz made in Section D.1. Appendix D: Fields inside a Round Wire 287 D.1 Partial Wave Expansion Warning : In this appendix, we use the same function name E to represent three different functions, E(r,θ,z,t) E(r,θ) E(r,m) The functions are distinguished by the arguments shown, and if they are not show n, the general context of the discussion will indicate which function is implied. The symbol E is thus "overloaded". (a) The General Method The starting point for the calculation is the damped wave equation (1.3.36, region 2) for the E field inside the wire. Unsubscripted parameters refer to properties of the wire. ( ∇ 2 - με ∂t2 - μσ∂t)E(r,θ,z , t ) = 0 . (1.3.36) Cylindrical coordinates (r, θ,z) are used, as appropriate for an infinite straight round wire. Recall that the damping term arises when the driving current J on the right of (1.2.1) is replaced by Ohm's Law J = σE. We now make the ansatz that a solution to th e above wave equation may be expressed in the following form where the t and z dependence is exposed and where E(r,θ) is a complex function to be determined: E(r,θ,z,t) = e j(ωt-kz) E(r,θ) . (D.1.1) The idea here is that we take our round wire to be one of two conductors of a transmission line (the other wire may or may not have a round cross section). The form shown in (D.1.1) says that the E field inside our round wire is assumed (an Ansatz!) to be a si mple "traveling wave" moving down this transmission line in the +z direction. As this interior wave mov es down the line, we expect to have an exterior wave whose E field takes the same general form shown in (D.1.1). If we match the E and B field boundary conditions of the interior and exterior waves, we e xpect k to have the same value on both sides of the round wire boundary. About k For a lossless wave, we expect the conductors to si mply deform the exterior fields (for example, causing the E field to be perpendicular to the conduc tor surfaces), but we expect the exterior wave to travel at the speed of light in th e dielectric, with no "drag" from th e conductors. In this lossless case, we then expect to have k = ω μdεd = ω/vd ≡ βd0 in (1.5.1b), where v d is the speed of light in the dielectric. If the dielectric conducts but the conductors are perfect, we have instead k = ω μdξd ≡ βd in (1.5.1a), and then k has a negative imaginary part which causes decay along the line, but k is still a characteristic of the dielectric medium. However, if the conductors are not perfect, then th ey too contribute to the decay, and in this case we expect that our parameter k will no longer be a characteri stic just of the dielectric. For example, we expect it will depend on R, the resistance per unit length of the conductors. Appendix D: Fields inside a Round Wire 288 We shall continue to use the generic parameter k throughout this appendix, to make sure our theory can handle situations with loss. One should think of k as a general complex parameter which (hopefully) has a negative imaginary part and whose real part is the wave phase velocity. Only in the special case of a completely lossless line do we have k = ω/vd = βd0. In the strong and extreme skin effect regimes, Chapter 4 develops a formula for the parameter k based on Maxwell's equations. This formula states that k = -j (R+jωL)(G+jωC) . This same formula arises in the network model of Appendix K, but in that model the formula applies all the way down to ω = 0. Probably this extrapolation of th e Maxwell-derived k expression down to low frequencies is reasonable though not exact. Having remarked on these two models for k, we shall ignore them until we reach Appendix D.11 and continue to work with our generic parameter k. When (D.1.1) is put into the above wave equa tion (1.3.36), time derivatives can be replaced ∂ t→ jω with the result ( ∇2 + β2) E(r,θ,z , t ) = 0 ( D . 1 . 2 a ) or [ ∇2D2 + (β2-k2) ] E(r,θ) = 0 ∇2 = ∇2D2 + ∂z2 (D.1.2b) where β 2 = μεω2 - jωμσ = ω2μ (ε - jσ/ω) = ω2μ ξ ξ ≡ ε - jσ/ω . (1.5.1c) We could have defined the temporal Fourier Transform of E(r,θ,z,t), E^(r,θ,z,ω') ≡ FT{ E(r,θ,z,t), ω'} = e -jkz E(r,θ) 2πδ(ω-ω') = e-jωt E(r,θ,z,t) 2πδ(ω-ω') as in (1.6.11) and then (D.1.2a) would be valid as well for E^(r,θ,z,ω) which would be a more conventional Helmholtz equation, but since E(r,θ,z,t) is monochromatic, we leave (D.1.2a) as is. One can regard (D.1.1) as an assumed variable-s eparated form for a solution, an "ansatz". If a consistent solution to the Maxwell equations can be found with this assumption, it is justified de facto. Sign Convention Comment : Section 1.6 discusses the Fourier Transform (1.6.8) where e+jωt appears in the expansion formula. For E^(x,ω) = 2πδ(ω-ω1) one gets E(x,t) = e+jω1t and then the form of a wave solution is e+j(ωt-kz) with the + sign associated with ωt. In general, EE people like to assume time dependence of the form e+jωt (and they prefer j in place of i for -1 ). The Fourier Transform is of course valid with the other sign choice for the two exponent ials, and for that other sign choice one would have E^(x,ω) = 2πδ(ω-ω1) => E(x ,t) = e-jω1t and one would think of a wave as e-j(ωt-kz) = e+j(kz-ωt). This sign convention is common in many physics texts [e.g. Jackson (7.8)], but in this document we use the e +j(ωt-kz) convention usually used in EE texts [e.g. Haus-Melcher 13.1 (7)]. Jackson suggests a physics/EE conversion algorithm of i ↔ -j. It is all just a convention choice and, as in (1.6.6), only the sign of the imaginary physical field under considera tion is affected. If one thinks of the physical field under consideration as Re{ E(x,t)}, the sign convention choice makes no difference at all. Appendix D: Fields inside a Round Wire 289 (b) Partial Wave Expansions The next step is to do a "partial wave expansion" (that is, a complex Fourier series expansion) of E(r,θ) in terms of "azimuthal harmonics" eimθ, so that the variable θ is replaced with the partial wave index m: E(r,θ) =∑ m = -∞∞ E(r,m) ejmθ // expansion (D.1.3a) E(r,m) = (1/2 π) ∫-π π dθ E(r,θ) e-jmθ . // projection (D.1.3b) In analogy with (D.1.1) we define a surface charge density n( θ,z,t) which has the following ansatz variable-separated form, n(θ,z,t) = e j(ωt-kz) n(θ) . ( D . 1 . 4 ) We then expand n( θ) as in (D.1.3), n(θ) = ∑ m = -∞∞ Nm ejmθ // Coul/m2 (D.1.5a) Nm = (1/2π) ∫-π π dθ n(θ) e-jmθ . // Coul/m2 ( D . 1 . 5 b ) Nm is the "moment" of the surface charge distribution in the mth partial wave. As with E(r,θ), the function n(θ) is also a function of implicit arguments ω and k. In principle, n( θ) could have a phase which varies with θ. If we momenarily assume this is not the case and assume that n( θ) is real, then (D.1.5b) says N -m = Nm* and then n(θ) = ∑ m = -∞∞ Nm ejmθ = N0 + ∑ m = 1∞ [ Nm ejmθ + Nm* e-jmθ ] = N0 + 2 ∑ m = 1∞ Re{ Nm ejmθ} = N 0 + 2 ∑ m = 1∞ { Re(N m) cos(mθ) - Im(N m) sin(mθ) } . (D.1.6) If we furthermore assume that n( θ) is an even function of θ, as symmetry implies for our particular figure below, then (D.1.5b) says the N m are real and then we have n(θ) = N0 + 2 ∑ m = 1∞ Nm cos(mθ) . // n( θ) real and even in θ (D.1.7) For a moderately closely spaced twin lead transmi ssion line (we allow for different radii), one might expect the m=0 and m=1 partial waves to be dominant : Appendix D: Fields inside a Round Wire 290 Fig D.1 Notice that N0 = (1/2π) ∫-π π dθ n(θ) = (1/2π) (1/a)(1/dz) ∫-π π [adθdz] n(θ) = (1/2π) (1/a)(1/dz) Q where Q is the total charge on a thin ribbon (width) dz wrapping the round wire. In (4.3.8) we refer to the quantity Q/dz as q(0), where q(z) = q(0) ejkz = the total charge on the wire per unit length. Thus, N0 = (1/2πa) q(0) = <n( θ) > ( D . 1 . 8 ) (c) The Vector Laplacian in Cylindrical Coordinates Given the foll owing cylindrical-coordinates field components, E(r,θ,z,t) = Er(r,θ,z,t)r^ + Eθ(r,θ,z,t)θ^ + Ez(r,θ,z,t)z^ we may write out our ansatz wave form (D.1.1) and the Helmholtz equation (D.1.2a) in more detail, Er(r,θ,z,t) = ej(ωt-kz) Er(r,θ) . [ ∇2E(r,θ,z,t)]r + β2 Er(r,θ,z,t) = 0 Eθ(r,θ,z,t) = ej(ωt-kz) Eθ(r,θ) . [ ∇2E(r,θ,z,t)]θ + β2 Eθ(r,θ,z,t) = 0 Ez(r,θ,z,t) = ej(ωt-kz) Ez(r,θ) . [ ∇2E(r,θ,z,t)]z + β2 Ez(r,θ,z,t) = 0 . (D.1.9) where β2 is the Helmholtz parameter of the conducto r medium, not to be confused with k. In Cartesian coordinates, it happens that [ ∇2E]i = ∇2(Ei), but this is not generally true for curvilinear coordinates. In cylindrical coordinates, it is true for the z coordinate only. The operator ∇2 when applied to a vector field is called "the vector Laplacian" and it is very different from the scalar Laplacian, so much so that some authors (Moon and Spencer) replace [ ∇ 2E] by [@E] which is defined in this manner [@E] ≡ [∇ 2E] ≡ grad(div E) – curl (curl E) = ∇(∇•E) - ∇ x (∇ x E) (D.1.10) whereas ∇ 2φ ≡ div(grad φ) = ∇ •(∇φ) . ( D . 1 . 1 1 ) Appendix D: Fields inside a Round Wire 291 It is the vector Laplacian that appears in our Helmholtz e quation (D.1.2). For cylindrical coordinates it turns out that, ( ∇ 2E)r = ∇2Er - (2/r2) ∂θEθ - (1/r2) Er ( ∇2E)θ = ∇2Eθ + (2/r2) ∂θEr - (1/r2) Eθ ( ∇2E)z = ∇2Ez ( D . 1 . 1 2 ) where ∇ 2 is the scalar Laplacian, given in cylindrical coordinates by ∇ 2 = (1/r)∂ r(r∂r) + (1/r2)∂θ2 + ∂z2 = ∂r2 + (1/r)∂ r + (1/r2)∂θ2 + ∂z2 . (D.1.13) Notice in (D.1.12) that E θ is mixed into the "r equation" and E r is mixed into the " θ equation". See for example Morse and Feshbach Vol I p 116, Moon and Spencer p 139, or do a web search on "vector Laplacian". The author's Tensor Analysis document, Sections 13, 14 and 15, derives these results for arbitrary coordinate systems. Here is a summary of vector differential operators in cylindrical coordinates taken from Morse and Feshbach, where the last line corresponds to the above discussion: (D.1.14) We use θ for azimuth instead of their φ since φ is our scalar potential. Using the ansatz form (D.1.1) and partial wave expa nsions of the form (D.1.3) or (D.1.5), it is a simple matter to convert an equation containing the above differential operators and involving components like E i(r,θ,z,t) or n(θ,z,t) to a simpler equation involving components like E i(r,m) and N m and this will be done below. (d) The three Helmholtz equations and div E = 0 (in partial waves) 1. The z equation: The Ez Helmholtz Equation from (D.1.9) is [ ∇2E]z + β2 Ez = 0. Using (D.1.12) and (D.1.13), the E z equation may be written, [∂r2 + (1/r) ∂r + (1/r2) ∂θ2 + ∂z2 + β2 ] ej(ωt-kz)Ez(r,θ) = 0 . (1) Appendix D: Fields inside a Round Wire 292 Inserting the expansion (D.1.3) for E z(r,θ) and moving the m sum to the left gives ∑ m = -∞∞ [∂r2 + (1/r) ∂r + (1/r2) ∂θ2 + ∂z2 + β2 ] ej(ωt-kz)Ez(r,m) ejmθ = 0 . (2) We can then make the obvious replacements ∂z = -jk and ∂θ = +jm to get, ∑ m = -∞∞ { [∂r2 + (1/r) ∂ r -m2 (1/r2) – k2 + β2 ] ej(ωt-kz)Ez(r,m) } ejmθ = 0 . (3) Due to the completeness of functions ejmθ on the interval (- π.π), we conclude that { } = 0, or [∂ r2 + (1/r) ∂r -m2 (1/r2) – k2 + β2 ] ej(ωt-kz)Ez(r,m) = 0 . (4) Alternatively one can apply ∫-π π dθ e-jm'θ to both sides of (3), use the orthogonality property ∫-π π dθ ej(m-m') θ = 2π δm,m' , ( 5 ) and then change m' to m to get (4). Next, multiply both sides of (4) by r 2 e-j(ωt-kz) to get, [r2∂r2 + r ∂r - m2 +r2( β2- k2)] Ez(r,m) = 0 . (D.1.15) We may then write these rules for conver ting equation (1) to equation (D.1.15) Conversion Rules : ∂z → -jk ∂t→ +jω ∂θ → +jm f(r,θ ,z,t ) → f(r,m) (D.1.16) We can now practice with these rules to convert various other equations of interest. A field with unstated arguments has the full arguments (r, θ,z,t). 2. The r equation: The E r Helmholtz Equation from (D.1.9) is [ ∇2E]r + β2 Er = 0 . Using (D.1.12) we find, ∇ 2(Er) - (2/r2) ∂θEθ - (1/r2) Er + β2Er = 0 [∂r2 + (1/r)∂ r + (1/r2)∂θ2 + ∂z2] Er - (2/r2) ∂θEθ - (1/r2) Er + β2Er = 0 [∂ r2 + (1/r)∂ r + (1/r2)∂θ2 + ∂z2 - (1/r2) + β2] Er - (2/r2) ∂θEθ = 0 . Appendix D: Fields inside a Round Wire 293 Now apply the conversion rules to get [∂ r2 + (1/r)∂ r + (1/r2) (-m2) - k2 - (1/r2) + β2] Er(r,m) - (2/r2) jm Eθ(r,m) = 0 . Group like terms and multiply by r2 to get [r2∂r2 + r∂r - (m2+1) + r2(β2-k2)] Er(r,m) - 2jm E θ(r,m) = 0 . (D.1.17) 3. The θ equation: The E θ Helmholtz Equation from (D.1.9) is [ ∇2E]θ + β2 Eθ = 0 . Using (D.1.12) we find, ∇ 2(Eθ) + (2/r2) ∂θEr - (1/r2) Eθ + β2Eθ = 0 [∂r2 + (1/r)∂ r + (1/r2)∂θ2 + ∂z2] Eθ + (2/r2) ∂θEr - (1/r2) Eθ + β2Eθ = 0 [∂ r2 + (1/r)∂ r + (1/r2)∂θ2 + ∂z2 - (1/r2) + β2] Eθ + (2/r2) ∂θEr = 0 . Now apply the conversion rules to get [∂ r2 + (1/r)∂ r + (1/r2)(-m2) + (-k2) - (1/r2) + β2] Eθ(r,m) + (2/r2) jm Er(r,m) = 0 . Group like terms and multiply by r 2 to get [r2∂r2 + r∂r - (m2+1) + r2(β2-k2)] Eθ(r,m) + 2jmE r(r,m) = 0 . (D.1.18) 4. The divE = 0 equation: Using (D.1.14) for div E ( times r) we write div E = 0 as ∂ r (r Er) + ∂θEθ + r ∂zEz = 0 . Applying the conversion rules gives ∂ r [r Er(r,m)] + jmE θ(r,m) + r (-jk)E z(r,m) = 0 or [1 + r ∂ r ] Er(r,m) + jmE θ(r,m) + r (-jk)E z(r,m) = 0 . (D.1.19) Here then is a summary of the above four results: Appendix D: Fields inside a Round Wire 294 The Three Helmholtz Equations and the div E = 0 equation (in partial waves) (D.1.20) [ ∇2E]z + β2 Ez = 0 : [ r2∂r2 + r ∂r - m2 + r2 ( β2- k2)] Ez(r,m) = 0 (D.1.15) [ ∇2E]r + β2 Er = 0 : [ r2∂r2 + r∂r - (m2+1) + r2(β2-k2)] Er(r,m) - 2jm E θ(r,m) = 0 (D.1.17) [ ∇2E]θ + β2 Eθ = 0 : [ r2∂r2 + r∂r - (m2+1) + r2(β2-k2)] Eθ(r,m) + 2jmE r(r,m) = 0 (D.1.18) div E = 0 : ∂ r [r Er(r,m)] + jmE θ(r,m) -jk r E z(r,m) = 0 (D.1.19) Helmholtz Comments : The scalar Helmholtz equation ( ∇2+β2)u(r,θ ,z) = 0 is fully separable in cylindrical coordinates and the "harmonics" (we call them atomic forms) are as follows [ J m(β'r), Ym(β'r)] * [ejmθ, e-jmθ] * [e jkz , e- jkz] ( D . 1 . 2 1 ) where k is a free real parameter and where β' 2 = β2 - k2. Here we use parameter names relevant for our particular problem where u = E z. These atomic forms appear for example in Moon and Spencer p15 with β = κ, m = p, β ' = iq, α 2 = m2, and -α3 = β'2. Whether a parameter like m or β' is real, imaginary or complex depends on the nature of the problem, and the above is a standard atoms choice for problems of our type. The θ "quantum number" m is quantized to be an in teger by the fact that our problem region is the entire range (- π,π) for θ and the solution must be single valued in θ. Our k is a parameter determined for a lossless line by k = ω μdεd and is thus correlated with the selected frequency ω, whereas our parameter β is always complex as in (1.5.1c or d). In gene ral, in any list of atomic forms like that shown above, two of the three atoms will be oscillatory and the third will be exponential, and in our case J m(β'r) is the exponential one, hence the skin effect with its exponential damping as shown in (2.3.7). Away from a singular point, any solution to ( ∇2 + β2) u(r,θ,z) = 0 must be writable as a linear combination of the atoms, so u = ∫dk Σm [Ak,m Jm(β'r) + Bk,m Ym(β'r) [Ck,m ejmθ + Dk,m e-jmθ ] [Ek,m e jkz + Fk,m e- jkz ]. A general solution method is to find a subset of the a bove most-general form that is appropriate in each "region" of the problem, and then to match boundary conditions between regions . If the problem is well- posed, this will determine all the constants A,B,C,D,E,F. We refer to this solution method as "the method of Smythian forms" (Smythe used this method a lot). Often many of these constants are 0. In contrast, the vector Helmholtz equation is not separable in cylindrical coordinates (see Moon and Spencer p 139), it is not even "R-separable", so there are no associated "harmoni cs" as there are with the Appendix D: Fields inside a Round Wire 295 scalar Helmholtz equation. Nevertheless, the functions ejmθ form a complete set for θ in (-π,π) and our expansion of each E i onto these ejmθ is certainly allowed, even though these ejmθ are not part of any associated harmonics for the vector Helmholtz equation. However, in Cartesian coordinates each Helmholtz component equation is a scalar Helmholtz equation. In cylindrical coordinates z is a Cartesian coordinate, so we should not be surprised when we find below that E z ~ Jm(β'r) eimθ e- jkz and this fits into the gene ral form noted above. Neither E r nor Eθ will have such a form. D.2 Solutions for E z,Er and E θ (a) The E z Solution As shown in (D.1.15), the Helmholtz equation for E z(r,m) is [r2∂r2 + r ∂r + (r2 β'2 - m2)] Ez(r,m) = 0 (D.2.1) where β' 2 = β2 - k2 . ( D . 2 . 2 ) For a perfect conductor, | β| is very large compared to the low-loss value k = β d ≈ βd0 (slight dielectric conductivity), so we could ignore the distinction between β and β' in that low-loss case. To show this, recall from (1.5.1b and d) that βd02 = ω2μdεd β2 ≈ - jωμσ => | β2 βd02 | ≈ μ μd σ ωεd . In scale, μ and μd are about the same, so using numbers from (1.1.28) and (1.1.29), | β2 βd02 | ≈ σ ωεd = 5.81 x 107 2πf 8.85 x 10-12 ≈ 1018 f(Hz) = 109 f(GHz) | β βd0 | ≈ 3.2x104 f(GHz) For f = 100 GHz we then find that | β/βd0| ≈ 3200, so for f < 100 GHz, | β/βd0| > 3200. As noted earlier, we maintain k as a general comple x parameter to be able to handle situations with loss, and thus we maintain the distinction between β and β' in all that follows. Setting x = β'r one finds ∂ r = β'∂x and then r ∂r = x∂x and so on so that (D.2.1) reads [x 2∂x2 + x ∂x + (x2- m2)] Ez(x/β',m) = 0 . x = β'r (D.2.3) Appendix D: Fields inside a Round Wire 296 This is Bessel's equation [ Spiegel 24.1] a nd the solution subject to the condition that E z be finite at r = 0 is Ez(x/β',m) = C zm Jm(x) or Ez(r,m) = C zm Jm(β' r ) ( D . 2 . 4 ) where C zm is an arbitrary constant for each partial wave m. For m = 0, equation (D.2.4) is consistent with (2.2 .22) found by other means. In Section 2.1 we dealt only with the m=0 partial wave, which embodies the symmetrical part of the problem. (b) The E r Solution As shown in (D.1.17), the Helmholtz equation for E r(r,m) is, using (D.2.2), [r 2∂r2 + r∂r - (m2+1) + r2β'2] Er(r,m) - 2jm E θ(r,m) = 0 (D.2.5) while the div E = 0 condition was stated in (D.1.19) as [1 + r ∂r ] Er(r,m) + jmE θ(r,m) + r (-jk)E z(r,m) = 0 or -jmE θ(r,m) = [1 + r ∂r ] Er(r,m) - r (jk)E z(r,m) . (D.2.6) Inserting this into (D.2.5) gives [r 2∂r2 + r∂r - (m2+1) + r2 β'2] Er(r,m) + [2 + 2r ∂r ] Er(r,m) - 2r (jk)E z(r,m) = 0 or [r2∂r2 + 3r∂r + (1-m2) + r2 β'2] Er(r,m) = 2r (jk)E z(r,m) . (D.2.7) Inserting solution (D.2.4) for E z(r,m) this becomes [r 2∂r2 + 3r∂r + (1-m2) + r2 β'2] Er(r,m) = 2r (jk) C zm Jm(β'r) or [r 2∂r2 + 3r∂r + (1-m2) + r2 β'2] Er(r,m) = 2j (k/ β') Czm β' r Jm(β'r) or [r 2∂r2 + 3r∂r + (1-m2) + r2 β'2] Er(r,m) = K m β'r Jm(β'r) (D.2.8) where K m ≡ 2j (k/β') Czm . ( D . 2 . 9 ) In order to get the left side of (D.2.8) into something recognizable, we define E r(r,m) = x-1 Fm( x ) ( D . 2 . 1 0 ) where x is a dimensionless radial variable whic h will play a major role in the following, Appendix D: Fields inside a Round Wire 297 x ≡ β'r and x a ≡ β' a . ( D . 2 . 1 1 ) Then (D.2.8) becomes [x 2∂x2 + 3x∂x + (1-m2) + x2] { x-1 Fm(x)} = 2j (k/β ') Czm x Jm(x) or x [x 2∂x2 + 3x∂x + (1-m2) + x2] { x-1 Fm(x)} = K m x2 Jm(x) . (D.2.12) Ever eager, Maple expands the left side of (D.2.12), so that (D.2.12) becomes [ x 2 ∂x2 + x ∂x + (x2-m2)] Fm(x) = Km x2 Jm(x) . (D.2.13) The left side of (D.2.13) is the normal Bessel operator [ Spiegel 24.1] , but the equation is also driven by a power times a Bessel function. The solution to the equation is the homogeneous solution of the Bessel equation plus the particular solution which is the r esponse to the driving function on the right hand side. The homogeneous solution is the usual linear combination of J m(x) and Y m(x), but we must reject Ym(x) since it blows up at x=0 and thereby causes the field E r to be singular, which it cannot be, smack in the middle of a wire. The particular solution is not very obvious a nd required some hunting to find. It is this F m(x)particular = (1/2) K m [ x Jm+1(x) ] . (D.2.14) as Maple confirms, continuing the above code, Therefore, we now have this full solution for F m(x) F m(x) = Fm(x)particular + Fm(x)homogeneous = (1/2) K m [ x Jm+1(x) ] + am Jm(x) and then from (D.2.10) the full solution for E r , Appendix D: Fields inside a Round Wire 298 Er(r,m) = a m x-1 Jm(x) + Km 2 Jm+1(x) . (D.2.15) For each value of m, there are two as-yet undetermined constants, a m and (Km/2). However, looking at (D.2.15), we see that, since J 0(x) ≈ 1 for small x, we must have a 0 = 0 ( D . 2 . 1 6 ) to keep E r finite at r = 0. Later we shall obtain expressions for a m and (Km/2). (c) The E θ Solution Recall (D.2.6) in slightly altered form, jmE θ(r,m) = - ∂r[rEr(r,m)] + r (jk)E z(r,m) . (D.2.6) We can then insert our known E z and Er to get E θ : Ez(r,m) = C zm Jm( x ) (D.2.4) Er(r,m) = a m x-1 Jm(x) + Km 2 Jm+1( x ) . (D.2.15) so (D.2.6) just above becomes the following : jmE θ(r,m) = - ∂r[r{ am x-1 Jm(x) + Km 2 Jm+1(x)}] + r (jk) C zm Jm(x) jmEθ(r,m) = - ∂x[x{ am x-1 Jm(x) + Km 2 Jm+1(x)}] + Km 2 x Jm(x) // K m ≡ 2j (k/β') Czm jmEθ(r,m) = - ∂x[am Jm(x) + Km 2 x Jm+1(x)] + Km 2 x Jm(x) jmE θ(r,m) = -a m Jm'(x) - Km 2 Jm+1(x) -Km 2 x Jm+1'(x) + Km 2 x Jm(x) jmE θ(r,m) = - a m Jm'(x) + Km 2 [ - Jm+1(x) - x J m+1'(x) + x J m(x) ] . (D.2.17) At this point we invoke the recurrence relations [ NIST 10.6.2 ], where C is any Bessel function, to write J m+1' = Jm - (m+1)x-1Jm+1 first relation with ν = m+1 Jm' = -Jm+1 + (m/x)J m second relation with ν = m . (D.2.18) Appendix D: Fields inside a Round Wire 299 Insert these into (D.2.17) to get jmE θ(r,m) = - a m Jm' + Km 2 [ - x Jm+1' - Jm+1 + xJm] = - a m {-Jm+1 + (m/x)J m } + Km 2 [ - x { J m - (m+1)x-1Jm+1} - Jm+1 + xJm] = a m Jm+1 - am (m/x)Jm + Km 2 [ - x Jm + (m+1) J m+1 - Jm+1 + xJm] = a m Jm+1 - am (m/x)Jm + Km 2 [ m Jm+1] = - a m (m/x)Jm + ( Km 2 m + a m ) Jm+1 . Dividing by m then gives the final solution, jEθ(r,m) = - a m x-1 Jm(x) + ( Km 2 + am m ) Jm+1(x) x = β'r . (D.2.19) We now gather up the solutions deve loped above, but first, recall that Km ≡ 2j (k/β') Czm (D.2.5) which we can solve to get C zm = (1/2j)(β '/k) Km . ( D . 2 . 2 0 ) Installing this into (D.2.4), our three E field components are then First summary of the E field solutions (D.2.21) Ez(r,m) = - j ( β'/k) Km 2 Jm(x) x = β'r (D.1.27) Er(r,m) = a m x-1 Jm(x) + Km 2 Jm+1(x) β'2 = β2 - k2 (D.2.11) jEθ(r,m) = - a m x-1 Jm(x) + ( Km 2 + am m ) Jm+1(x) (D.2.15) These solutions were obtained from the z a nd r Helmholtz equations and from the div E = 0 equation. It is an easy matter to have Maple verify that this solution set solves the div E equation and all three of the Helmholtz equations z, r and θ : [ here k is called bd ] Appendix D: Fields inside a Round Wire 300 In Maple one must be careful with this kind of verification to make sure Maple has not misunderstood something. For example, perhaps it thinks ∂rEz = 0 because it thinks E z is a constant. This is the purpose of using the "inert" Diff operators (versus diff) and then forcing them to evaluate later with the value() operator. One should always view expressions before si mplification to make sure things are kosher. For example, changing the colon to semicolon af ter value(e1) to force display, one gets Appendix D: Fields inside a Round Wire 301 Here Maple has duly computed the Bessel function derivatives in expression e1 but does not yet realize that the expression is 0. This is brought out by the simplify(%) command (simplify that last computed expression) and the output of the simplify command is the 0 on the last line. (d) The Charge Pumping Boundary Condition The reason w e are interested in the surface charge n( θ) of (D.1.5) is that it acts as a driving source of the radial electric field inside the wire. Recall the e quation of continuity (1.1.35) converted to the ω domain div J = - jωρ ⇔ -jω [∫V ρ dV] = ∫S J • dS . (D.2.22) This is meant to be (1.1.25) where J is conduction current and ρ is free charge. When applied to a thin box of radial area dS straddling the wire surface, Fig D.2 one finds that ∫S J • dS = -Jr(r=a-ε,θ)dS and ∫V ρ dV = n(θ) dS so that ( ε implies just below surface) Jr(r=a-ε,θ) = jω n(θ) . ( D . 2 . 2 3 ) We assume that there is no conduction current outside the wire to get this result (non-conducting dielectric). Since J = σE, this is really a boundary condition on the radial electric field just below the surface, E r(r=a-ε,θ) = (jω/σ) n(θ) . ( D . 2 . 2 4 ) We convert this to m-space using the convers ion rules (D.1.16) to obtain (dropping the ε) Appendix D: Fields inside a Round Wire 302 Er(r=a,m) = (j ω/σ) Nm . ( D . 2 . 2 5 ) Thus, the interior radial electric field must have a certain value at the r=a boundary in each partial wave, and this value is determined by the mome nt of the surface charge distribution. By way of interpretation, the surface charge of a transm ission line is "pumped" by the radial current in the wire (in quadrature). This radial current is accompan ied by the usual longitudinal current one expects to find inside the conductors of a transmission line. If the dielectric conducts with some σ d > 0 but σd << σ , one must make these replacements in (D.2.24) and (D.2.25), n(θ) → (ξd/εd) n(θ) N m → (ξd/εd) Nm . See (D.9.23) and surrounding discussion. Generally we shall assume σ d = 0 in the following work just to avoid having the extra ( ξd/εd) factors floating around. (e) Application of the Boundary Conditions Our task here is to derive expressions for the constants a m and Km appearing in the above E field component equations. We have two boundary conditions to impose: Er(r=a,m) = (j ω/σ) Nm ( D . 2 . 2 6 ) Eθ( r = a , m ) = 0 ( D . 2 . 2 7 ) The first is the radial charge pumping condition s hown in (D.2.25) above. Th e second boundary condition is an assumption that requires its own discussion in Section D.8 below. It implies that the cross sectional wire surface is an equipotential surface and that therefore E θ(r=a,θ) = 0. This in turn requires that in each partial wave E θ(r,m) = 0 since Eθ(r,m) = (1/2 π) ∫-π π dθ Eθ(r,θ) e-jmθ (D.1.3b) Eθ(a,m) = (1/2 π) ∫-π π dθ Eθ(a,θ) e-jmθ = (1/2π) ∫-π π dθ 0 e-jmθ = 0 . These two boundary conditions serve to determine the two constants a m and Km, though a bit of algebra is required. The first step is to use the E r and Eθ expressions shown in summary box (D.2.21) to write out the two boundary conditions as am xa-1 Jm(xa) + Km 2 Jm+1(xa) = (jω /σ) Nm (1) - am xa-1 Jm(xa) + ( Km 2 + am m ) Jm+1(xa) = 0 . (2) Appendix D: Fields inside a Round Wire 303 Addition and subtraction of these e quations gives two new equations, Km 2 Jm+1(xa) + ( Km 2 + am m ) Jm+1(xa) = (jω/σ) Nm (3) 2 am xa-1 Jm(xa) - am m Jm+1(xa) = (jω /σ) Nm . (4) Using the recursion relation 2m x-1 Jm = [Jm+1 + Jm-1] , the second may be immediately solved for a m, am m = (jω/2σ) 2 Nm 1 Jm-1(xa) . ( 5 ) Using this same recursion relation and (5) for a m , equation (2) may be solved to get ( Km 2 + am m ) = (jω/2σ) Nm [1 Jm+1(xa) + 1 Jm-1(xa) ] . (6) Finally, subtracting (5) from (6) we find K m 2 = (jω/2σ) Nm [1 Jm+1(xa) – 1 Jm-1(xa) ] . (7) Notice the following situations for m = 0, a0 = 0 (8) // from (5) a0 0 = (jω/2σ) 2 N0 1 J-1(xa) = - (jω/2σ) 2 N0 1 J1(xa) (9) // from (5) K0 2 = (jω/2σ) N0 [1 J1(xa) – 1 J-1(xa) ] = ( jω/σ) N0 1 J1(xa) (10) // from (7) (K0 2 + a0 0 ) = 0 (11) // adding (9) and (10) We summarize the coefficients as follows: a m = (jω/2σ) 2m Nm 1 Jm-1(xa) . a 0 = 0 (D.2.28) Km 2 = (jω/2σ) Nm [1 Jm+1(xa) – 1 Jm-1(xa) ] K0 2 = (jω/σ) N0 1 J1(xa) (Km 2 + am m ) = (jω/2σ) Nm [1 Jm+1(xa) + 1 Jm-1(xa) ] (K0 2 + a0 0 ) = 0 Appendix D: Fields inside a Round Wire 304 The third equation is obvious from adding the first two, and Maple verifies that the first two satisfy (1) and (2). At this point it is convenient to introduce th e DC resistance per unit leng th of the wire (C.1.1), Rdc = 1 σπa2 ( D . 2 . 2 9 ) along with a new symbol to indicate the relative surface charge moment, ηm ≡ Nm N0 . ( D . 2 . 3 0 ) The DC moment N 0 can be related to the total cu rrent I in the wire as follows: I = ∫0 2π dθ ∫0 a r dr Jz(r,θ) = ∫0 2π dθ ∫0 a r dr { σ ∑ m = -∞∞ Ez(r,m) ejmθ } // (D.1.3a) = σ ∑ m = -∞∞ ∫0 a r dr Ez(r,m) ∫0 2π dθ ejmθ = 2π σ ∫0 a r dr Ez(r,0) = 2 π σ ∫0 a r dr {-j(β'/k) K0 2J0(x) } // (D.2.21) for E z(r,0) = - j ( β'/k) 2π σ K0 2 ∫0 a r dr J0( x ) / / x = β'r so xdx = β'2 rdr = - j ( β'k)-1 2πσ K0 2 [ ∫0 xa dx x J0(x)] = -j(β 'k)-1 2πσ K0 2 [ xa J1(xa) ] // GR7 5.52.1 = - j ( β'k)-1 2πσ {(jω/σ) N0 / J1(xa)} [ xa J1(xa) ] // (D.2.28) for K0 2 = ( β'k)-1 2πω N0 xa = (β'k)-1 2πω N0 β'a = 2 πω (a/k) N 0 so that N0 = (k/2πωa) I . // I is called i(z=0) in (4.9.2) so I = i(0) (D.2.31) As a check on this last result, recall from (D.1.8) that N 0 = (1/2πa) q(0), so (D.2.31) says (1/2πa) q(0) = (k/2 πωa) i(0) or q(0) = (k/ω ) i(0) . Given q(z) = q(0) e -jkz as in (4.3.8), i(z) = i(0) e-jkz as in (4.9.2), and k = ω/vd as in (2.1.4b) (for a lossless line) we obtain q(z) = i(z) /v d . // as in (4.11.19a) Appendix D: Fields inside a Round Wire 305 It follows from (D.2.31) that the normalization factor appearing in (D.2.28) may be written as (jω /2σ) Nm = (jω/2σ) Nm N0 N0 = (jω/2σ) ηm [(k/2πωa) I ] = (j/4) ηm (ak/σπa2) I = (j/4) (ak) ηm I Rdc . ( D . 2 . 3 2 ) We may now construct the final form for our E field solutions in (D.2.21) using the coefficients in (D.2.28) and the replacement (D.2.32) : E z(r,m) = -j( β'/k) Km 2 Jm(x) = -j( β'/k) (jω/2σ) Nm [1 Jm+1(xa) – 1 Jm-1(xa) ] Jm(x) = -j( β'/k) [(j/4) (ak) ηm I Rdc] [1 Jm+1(xa) – 1 Jm-1(xa) ] Jm(x) = (1/4) ηm I Rdc (aβ') [ Jm(x) Jm+1(xa) - Jm(x) Jm-1(xa) ] Er(r,m) = a m x-1 Jm(x) + Km 2 Jm+1(x) = [ ( j ω/2σ) Nm] { 2m 1 Jm-1(xa) x-1 Jm(x) + [1 Jm+1(xa) – 1 Jm-1(xa) ] Jm+1(x) } = (j/4) (ak) ηm I Rdc { 2mx-1Jm(x) Jm-1(xa) + Jm+1(x) Jm+1(xa) - Jm+1(x) Jm-1(xa) } = (j/4) (ak) η m I Rdc { Jm+1(x) Jm+1(xa) + Jm-1(x) Jm-1(xa) } where in the last line we used the NIST (10.6.1) Bessel identity (2m/x)J m(x) = Jm-1(x) + Jm+1(x). Next, jEθ(r,m) = - a m x-1 Jm(x) + (Km 2 + am m ) Jm+1(x) = [ ( j ω/2σ) Nm] { - 2m x-1 1 Jm-1(xa) + [ 1 Jm+1(xa) + 1 Jm-1(xa) ] Jm+1(x) = (j/4) (ak) ηm I Rdc { - 2mx-1Jm(x) Jm-1(xa) + [ Jm+1(x) Jm+1(xa) + Jm+1(x) Jm-1(xa) ] } = (j/4) (ak) η m I Rdc { Jm+1(x) Jm+1(xa) - Jm-1(x) Jm-1(xa) } Appendix D: Fields inside a Round Wire 306 Gathering up one more time: Second summary of the E field solutions : R dc = 1 σπa2 β'2 = β2 - k2 (D.2.33) Ez(r,m) = (1/4) η m I Rdc (aβ') fm f m = [ Jm(x) Jm+1(xa) - Jm(x) Jm-1(xa) ] x = β 'r Er(r,m) = (j/4) η m I Rdc (ak) gm g m = [ Jm+1(x) Jm+1(xa) + Jm-1(x) Jm-1(xa) ] x a = β'a Eθ(r,m) = (1/4) ηm I Rdc (ak) hm h m = [ Jm+1(x) Jm+1(xa) - Jm-1(x) Jm-1(xa) ] Maple verification of these solutions is shown below. Observations about the solution: (1) We looked for a traveling wave solution inside a round wire in which phase fronts propagate down the wire (z direction) with angular frequency ω and wavelength λ = 2π/Re(k). We found the solution shown in the above box. This solution satisfies all three co mponents of the vector Helmholtz equation (D.1.2) as well as the div E = 0 equation. (2) For a low-loss line one has ξ ≈ σ/(jω) and ξ d ≈ εd. These are the complex dielectric "constants". The corresponding wavenumbers are then β' ≈ β = ω μξ ≈ ω μ σ/(jω) = ej3π/4 ωμσ = ej3π/4 (2 /δ) (1.5.1c) ,(2.2.19), (2.2.21) k = β d = ω μdξd ≈ ω μdεd = ω / vd = βd0 vd = speed of light in the dielectric (D.2.34) Thus, in our wave solution (D.1.1), the phase fronts propagate down the inside of the wire at v d, the speed of light in the dielectric outside the wire. Although we have been quiet about the fields outside the wire, it seems reasonable to presume there is a wave outside also moving down the wire at v d . See Section D.7. (3) For r near a, where most of the action occurs due do the skin effect, the Bessel function ratios appearing in (D.2.33) are on the general order of unity so we expect the three brackets [...] to be of the same general size. It then follows that the E r and Eθ fields are smaller than E z by the ratio |k/ β'| which we have shown in the discussion below (D.2.2) is very small at frequencies below 100 GHz (lossless). Since Ez is an electric field inside copper, it is already itself quite small, so the E r and Eθ fields are extremely small. This then justifies their omission from the development of Chapter 2. (4) If there exist moments N m of the surface charge distribution on the wire with m > 1, then the corresponding ηm ≠ 0 and it is clear that E z(r,θ) and hence J z(r,θ) are non-uniform inside the wire. That Appendix D: Fields inside a Round Wire 307 is, these fields vary with θ as cos(m θ) as well as with r. The non-uniformity is not "small" but has the full strength of ηm. Of course we only expect to get significant moments of charge density n(θ ) when conductors are "fat and close". See (6.5.4) for the sp ecial case of both conductors being round wires, and then Section 6 (b) for more on this "proximity effect". (5) The surface impedance from (C.2.1) is just Z s(θ) = Ez(r=a,θ)/I. Thus, from (D.1.3a), Zs(θ) = (1/I) ∑ m = -∞∞ Ez(a,m) ejmθ / / (D.1.3a) = (1/4) R dc ∑ m = -∞∞ ηm [ xaJm(xa) Jm+1(xa) - xaJm(xa) Jm-1(xa) ] ejmθ // (D.2.33) w h e r e , ( D . 2 . 3 5 ) ηm = Nm/N0 = ωa kI ∫-π π dθ n(θ) e-jmθ // (D.1.5b) and (D.2.31) Thus we see the expected non-uniformity of Z z(θ) around the perimeter of the wire cross section due to the m ≠ 0 surface charge components. Maple verification of box (D.2.33) We use the same method illustrated below box (D.2.21) . The same expressions e1,e2,e3,e4 are entered as the left sides of the four equations whose right sides we expect to be 0. Then: [ needs repair ] Appendix D: Fields inside a Round Wire 308 D.3 What about the E θ Helmholtz Equation ? A review of the above derivation of the three fields E z, Er and Eθ shows that the E θ Helmholtz equation has been completely ignored. The E θ expression was obtained from the div E = 0 equation after the E z and Er fields were computed. It is reasonable to wonder whether the solution fi elds we have found above in fact solve this θ Helmholtz equation which mixes the E r and Eθ fields together in a manner similar to the r Helmholtz equation. A related question is whether the three Helmholtz equations and div E = 0 are four independent equations, or is one of the three Helmholtz equations dependent? In Cartesian coordinates suppose we know that (implied sums on repeated indices) (∂ j∂j + β2) E1 = 0 (∂j∂j + β2) E2 = 0 ∂iEi = 0 . // div E = 0 ( D . 3 . 1 ) Can we show that ( ∂ j∂j + β2) E3 = 0 so this third Helmholtz equa tion is dependent? If we apply the operator (∂ j∂j + β2) to the last equation above we get (∂j∂j + β2) ∂iEi = 0 or ∂i (∂j∂j + β2) Ei = 0 or ∂ 1 (∂j∂j + β2) E1 + ∂2 (∂j∂j + β2) E2 + ∂3 (∂j∂j + β2) E3 = 0 or ∂ 3 [(∂j∂j + β2) E3] = 0 ( D . 3 . 2 ) Appendix D: Fields inside a Round Wire 309 This does not prove that ( ∂j∂j + β2) E3 = 0 since (∂ j∂j + β2)E3 = f(x1,x2) ≠ 0 also satisfies (D.3.2). Rather than pursue this question further, we simply note that the Maple code below box (D.2.21) verifies that the E field solutions given in that box do indeed satisfy the θ Helmholtz equation (as well as the other two Helmholtz equations and the div E = 0 equation). Reader Exercise: Come up with some reason that this had to be the case. D.4 Computation of the B fields in the round wire The B field com ponents may be computed from the Maxwell curl E equation (1.1.2) - ∂tB = curl E . Maxwell curl E equation (1.1.2) (D.4.1) In cylindrical coordinates one has from (D.1.14), curl E = r^ [ r -1∂θEz - ∂zEθ] + θ^ [∂zEr - ∂rEz] + z^ [ r-1∂r(rEθ) - r-1∂θEr ] (D.4.2) where the fields are of the traveling wave form shown in (D.1.1) which we assume also for the B field. Thus, combining (D.1.1) with (D.1.3a), one has E(r,θ,z,t) = ej(ωt-kz) E(r,θ) = ej(ωt-kz) ∑ m = -∞∞ E(r,m) ejmθ (D.4.3) B(r,θ,z,t) = ej(ωt-kz) B(r,θ) = ej(ωt-kz) ∑ m = -∞∞ B(r,m) ejmθ . (D.4.4) Inserting the three cylindrical components of the E expansion (D.4.3) into (D.4 .2), one finds that these replacements may be made, ∂ t → +jω ∂z → -jk ∂θ → + j m . ( D . 4 . 5 ) Similarly, inserting the B expansion (D.4.4) into - ∂ tB one may replace ∂ t→ +jω. After doing this, both sides of (D.4.1) are expansions havi ng the general form of (D.4.3) and one may then equate terms in the m sum [completeness of the ejmθ on (-π.π)] to find that -jω B(r,m) = r^ [ r-1jmEz +jkEθ] + θ^ [-jkEr - ∂rEz] + z^ [ r-1∂r(rEθ) - r-1jmEr ] (D.4.6) and this then gives the three components of the B field B r(r,m) = (j/ ω) [curl E] r = (j/ω) [r-1jmEz +jkEθ] Bθ(r,m) = (j/ ω) [curl E] θ = (j/ω)[-jkEr - ∂rEz] Bz(r,m) = (j/ ω) [curl E]= (j/ ω) [r-1∂r(rEθ) - r-1jmEr] . (D.4.7) Appendix D: Fields inside a Round Wire 310 It is now a mechanical task to insert our E field components, and such tasks are grist for Maple's mill. We use the E component forms summary box (D.2.21) which have the a m and Km constants not yet specified. The alias line "unaliases" I, sets j = -1 in place of the default I, and allows simple reference to the Bessel functions of interest. Diff(Ez,r) represents ∂rEz, but in an "inert" form which is not executed until later after E z has been specified. The resu lting B field expressions are somewhat ugly but can be cleaned up using a few more Maple manipulations. Having seen the results, we extract certain factors as shown in the following commands, which we then translate back into our normal notation, (ω/β')Bz(r,m) = (Km 2 + am m )Jm(x) (ω/jβ')Br(r,m) = + ( mKm 2 1 rk - k rβ'2 am) Jm(x) + k β' (Km 2 + am m ) Jm+1(x) (ω/β')Bθ(r,m) = - ( mKm 2 1 rk - k rβ'2 am) Jm(x) + Km 2 (k β' + β' k ) Jm+1(x) . (D.4.8) Appendix D: Fields inside a Round Wire 311 The last two equations contain the same term which can be written as (recall x = r β') ( mKm 2 1 rk - k rβ'2 am) = ( mKm 2 β' rβ'k - k rβ'2 am) = ( mKm 2 β' k - k β' am)(1/x) The three equations for the exact B field components in the round wire are then shown in the summary box below which includes the earlier E field results as well: Summary of E and B fields inside a round wire (D.4.9) Ez(r,m) = - j ( β'/k) Km 2 Jm(x) x = β'r β'2 = β2 - k2 Er(r,m) = a m x-1 Jm(x) + Km 2 Jm+1(x) . jEθ(r,m) = - a m x-1 Jm(x) + ( Km 2 + am m ) Jm+1(x) (D.2.21) Bz(r,m) = ( β'/ω) (Km 2 + am m )Jm(x) Br(r,m) = j( β'/ω){ + ( mKm 2 β' k - k β' am) x-1Jm(x) + k β' (Km 2 + am m ) Jm+1(x) } Bθ(r,m) = ( β'/ω){ - ( mKm 2 β' k - k β' am) x-1Jm(x) + Km 2 ( k β' + β' k ) Jm+1(x) } , (D.4.8) where the constants are given in (D.2.2 8), which we rewrite using (D.2.32), a m = (j/4) (ak) ηm I Rdc * 2m 1 Jm-1(xa) Km 2 = (j/4) (ak) ηm I Rdc * [1 Jm+1(xa) – 1 Jm-1(xa) ] (Km 2 + am m ) = (j/4) (ak) ηm I Rdc * [1 Jm+1(xa) + 1 Jm-1(xa) ] . The three constant quantities at the end of the above summary box are roughly the same size in terms of scale. Using this fact, and the fact that for a low-loss line | β'| >> |k| = | βd| (so β = β'2+k2 ≈ β') we can simplify (D.4.9) to read, Bz(r,m) = ( β/ω) (Km 2 + am m )Jm(x) β' ≈ β, x = βr Br(r,m) = j( β/ω){ + ( mKm 2 β k ) x-1Jm(x) } // m ≠ 0 Bθ(r,m) = ( β/ω){ - ( mKm 2 β k ) x-1Jm(x) + Km 2 (β k ) Jm+1(x) } . (D.4.10) Appendix D: Fields inside a Round Wire 312 The last line of (D.4.10) can be further simplified, B θ(r,m) = ( β/ω){ - ( mKm 2 β k) x-1Jm(x) + Km 2 (β k ) Jm+1(x) } = ( β/ω) Km 2 βk 12 { - 2mx-1Jm(x) + 2J m+1(x) } = ( β/ω) Km 2 βk 12 { - Jm+1(x) - Jm-1(x) + 2J m+1(x) } // Spiegel 24.17 identity = ( β/ω) Km 2 βk 12 { Jm+1(x) - Jm-1(x) } . (D.4.11) Therefore, in the limit | β'| >> |k| = | βd| equations (D.4.10) become Bz(r,m) = ( β/ω) (Km 2 + am m )Jm(x) β' ≈ β Br(r,m) = j( β/ω) β βd{ mKm 2 x-1Jm(x) } // m ≠ 0 Bθ(r,m) = ( β/ω) β βd Km 2 1 2 [ Jm+1(x) - Jm-1(x)] . (D.4.12) For m>0, |B r| and |B θ| are larger than |B z| by the large factor | β /βd|. For m = 0, βr ≈ 0 so |βθ| >> |Bz|. It is this large B θ field which appears in Chapter 2 as B θ. D.5 Verification that the E and B fields satisfy the Maxwell equations The Maple program discussed above goes on to verify that the exact E and B fields obtained above for the round wire in fact satisfy Maxwell's equations. Since the B equations were obtained from the curl E Maxwell equation, this one is not verified. The othe r three Maxwell equations are projected into their partial wave versions anal ogous to (D.4.6) above : div B(r,m) = r -1∂r(rBr) + r-1∂θBθ + ∂zBz = r-1∂r(rBr) +r-1jmBθ -jk Bz ( D . 5 . 1 ) div E(r,m) = r-1∂r(rEr) + r-1∂θEθ + ∂zEz = r-1∂r(rEr) + r-1jmEθ - jkEz ( D . 5 . 2 ) curl B(r,m) = r^ [ r-1∂θBz - ∂zBθ] + θ^ [∂zBr - ∂rBz] + z^ [ r-1∂r(rBθ) - r-1∂θBr ] = r^ [ r-1jmBz + jkBθ] + θ^ [-jkBr - ∂rBz] + z^ [ r-1∂r(rBθ) - r-1jmBr ] . (D.5.3) Inside the round wire we expect to find Appendix D: Fields inside a Round Wire 313 div B = 0 div E = 0 // no free charge curl B = μ J + μ jωεE = μ(σ + jωε ) E = μ(jω)( ε - jσ/ω) E = jω μξ E = j ( β2/ω) E . // see (1.5.1c) Thus, for the divergence equations we just compute the divergence as shown and see if it comes out zero, while for the curl B equation we verify that curl B - j(β 2/ω) E = 0 ( D . 5 . 4 ) for each component. The fact (D.2.2) that β' 2 = β2 - k2 is also used. Here then is the Maple code which does the verification of the three Maxwell equations: In Maple % refers to the last quantity computed. Pr ior to each simplify(%) statement we find a huge mess for the expression at hand, but simplify then shows it is really zero. As an example, here is the execution of the verification that [curl B] z - j(β2/ω) Ez = 0 : Appendix D: Fields inside a Round Wire 314 No approximations were made in the E fields, th e B fields, or in these Maxwell verifications. D.6 The exact E and B fields for the m=0 partial wave The m=0 partial wave is all there is for an axially symmetric problem like that considered in Chapter 2, where the round wire is imagined in isolation, but is operationally the central conductor of a coaxial cable with a very distant return cylinder (outer shield). Here is the reduction of box (D.4.9) for m = 0, making use of the m=0 coefficients noted in box (D.2.28), namely, m = 0 a 0 = 0 (K0 2 + a0 0 ) = 0 K0 2 = (jω/σ) N0 1 J1(xa) Summary of E and B fields inside a round wire ( m = 0 only ) (D.6.1) Ez(r,0) = - j (β '/k) K0 2 J0(x) // large x = β'r β '2 = β2 - k2 Er(r,0) = K0 2 J1(x) . // small K0 2 = (j/2) (ak) I R dc 1 J1(xa) jEθ( r , 0 ) = 0 R dc = 1 σπa2 Bz(r,0) = 0 Br(r,0) = 0 Bθ(r,0) = ( β'/ω) ( k β' + β' k ) K0 2 J1(x) // large ~ β' (β'/k) No approximations have been made in these results, but a very good approximation for a low loss line is that |β| >> |k| = |β d| which means β' ≈ β, as discussed below equation (D.2.2). With this approximation, we Appendix D: Fields inside a Round Wire 315 have commented in the above box on the size of the various field components. The dominant components are E z(r,0) = - j ( β/k) K0 2 J0(x) = - j (β /k) (j/2) (ak) I R dc J0(x) J1(xa) = (1/2) β a I Rdc J0(x) J1(xa) = (ω/β) (1/2) (aβ2/ω) I Rdc J0(x) J1(xa) B θ(r,0) = ( β/ω) (β k ) (j/2) (ak) I R dc J1(x) J1(xa) = (j/2) (a β2/ω) I Rdc J1(x) J1(xa) . As shown in (2.2.3) we can write β2/ω ≈ - jμσ so that (1/2) (a β2/ω) I Rdc = (1/2) a (- j μσ) I 1 σπa2 = - j μI 2πa and then the dominant components above become Ez(r,0) = (ω/β) [ - j μI 2πa ] J0(x) J1(xa) = -j (ω/β) μI 2πa J0(x) J1(xa) B θ(r,0) = j [- jμI 2πa ] J1(x) J1(xa) = μI 2πa J1(x) J1(xa) . (D.6.2) These results are in agreement with E(r) and B(r) s hown in summary box (2.2.30) from the Chapter 2 calculation where we assumed E = E(r) z^ and B = B(r) θ^ D.7 What about the E fields outside the round wire? The Helm holtz equation (D.1.2) outside the wire contains βd instead of β. If we assume perfect conductors, then k = βd as well in (D.1.1). This means that β'2 = βd2- βd2 = 0 . We can then translate box (D.1.20) by replacing β2- k2 → 0 and β2 → k2 = βd2 to get the following "exterior" versions: (Note that ∇2 = ∇2 2D + ∂z2) [∇2E]z + k2 Ez = 0 : / / [ ∇2 2DE]z = 0 [r2∂r2 + r ∂r - m2] Ez(r,m) = 0 (D.1.15) ext [∇ 2E]r + k2 Er = 0 : / / [ ∇2 2DE]r = 0 [r2∂r2 + r∂r - (m2+1)] Er(r,m) - 2jm E θ( r , m ) = 0 (D.1.17) ext Appendix D: Fields inside a Round Wire 316 [∇2E]θ + k2 Eθ = 0 : / / [ ∇2 2DE]θ = 0 [r2∂r2 + r∂r - (m2+1)] Eθ(r,m) + 2jmE r( r , m ) = 0 (D.1.18) ext div E = 0 : ∂r [r Er(r,m)] + jmE θ(r,m) -j k r E z( r , m ) = 0 (D.1.19) ext The differential operators appearing in the above e quations are no longer Bessel-style operators, they are Euler-style operators. Euler ODEs have the general form [ r 2∂r2 + a r ∂r + b] f(r) = 0, and the solutions have this form ( from p 45 of Polyanin's exce llent ODE compendium, or just use Maple), For Ez(r,m) the equation (D.1.15) ext shown just above is in fact an Euler equation which has a = 1 and b = -m2 so μ = m and the solution forms are these (r ≥a outside the wire), Ez(r,m) = A mrm + Bmr-m m > 0 Ez(r,0) = C zln(r) + D z m = 0 . ( D . 7 . 1 ) Since z is a Cartesian coordinate, [ ∇2 2DE]z = 0 is the same as ∇2 2DEz = 0 which is just the 2D Laplace equation. When this equation is solved in polar coordinates (r, θ), one finds E z = Ez(r,m) ejmθ and the expressions shown above are the standard atomic forms for the radial function. See for example Stakgold Vol II p 92 (6.7) and following discussion. We can mimic our interior solution method presented in Section D.2 above, using the div E = 0 equation to eliminate E θ, and eventually end up with expressions for the three field components outside the wire. For m > 1 the general form for the exterior solution is found to be, Ez(r,m) = A m rm + Bm r-m Er(r,m) = -(jk/2) 1 m-1 Bm r1-m - 2j k 1 m2-1 Am r1+m + Cm rm-1 + Dm r-m-1 jEθ(r,m) = (jk/2) 1 m-1 Bm r1-m + j k m2+ 2m+3 m(m2-1) Am r1+m - Cm rm-1 + Dm r-m-1 (D.7.2) Appendix D: Fields inside a Round Wire 317 where there are now four constants A m, Bm, Cm and Dm to be determined in each partial wave. One could match the three E-field boundary conditions at r = a as per (1.1.50) (subscript d means dielectric) Ez(a,m) = E zd(a,m) ξ Er(a,m) = ξd Erd(a,m) jEθ(a,m) = jE θd( a , m ) ( D . 7 . 3 ) using the interior solutions shown in (D.2.33) where E θ(a,m) = 0. This gives 3 conditions on the 4 unknown constants so these boundary conditions can be met. The problem with this exterior solution method is that more information is needed to solve the problem. The "Smythian Form" solution (D.7.2) is fine, but it only applies inside a thick cylindrical shell (blue) whose inner diameter is r = a and whose outer di ameter is r = b, where b causes this shell to touch the nearest other conductor, as illustrated here, Fig D.3 The reason is that the dielectric E-field wave (Helmholtz) equation is not valid inside the "other conductor", so the form (D.7.2) cannot apply in a regi on which includes any of this other conductor. Since the blue shell region does not include r = ∞, one cannot rule out coefficients like A m and Cm. One is now stuck worrying about boundary conditions at r = b and the whole problem becomes intractable. But if one could find the complete exact exterior solution, one would find that inside the blue cylindrical shell the solution's partial wave fields would have the form shown in (D.7.2). Reader Exercise: (a) Verify (D.7.2). (b) In Chapter 6 a transmission line with two round c onductors is solved "exactly". Convert the solution to a coordinate system like that shown above, compute the E i(r,m) using (D.1.3b), and verify that these Ei field components fit into the form shown in (D.7.2). D.8 About the boundary condition E θ(a,m) = 0 We start with a quick review. In earlier sections of th is Appendix we examined the electric field inside a round wire (radius a) which was regarded as a conductor in a straight transmission line. The electric field was assumed to have the form of a longit udinal wave traveling down the conductor, E(r,θ,z,t) = e j(ωt-kz) E(r,θ) , (D.1.1) Appendix D: Fields inside a Round Wire 318 where k is the wavenumber parameter of the surro unding dielectric medium. We expanded the function E(r,θ) onto azimuthal partial waves ejmθ and solved the Helmholtz wave equation inside the wire with solutions as shown in box (D.2.21), E z(r,m) = - j ( β'/k) Km 2 Jm(x) x = β' r (D.1.27) Er(r,m) = a m x-1 Jm(x) + Km 2 Jm+1( x ) . (D.2.11) jEθ(r,m) = - a m x-1 Jm(x) + ( Km 2 + am m ) Jm+1(x) . (D.2.15) where β'2 = β2-k2 with β being the (complex) wavenumber parameter of the conductor, and where a m and Km are undetermined constants. At this point we applied the two boundary conditions, assuming a non-conducting dielectric, Er(r=a,m) = (j ω/σ) Nm (D.2.26) Eθ( r = a , m ) = 0 . (D.2.27) where N m is the mth partial wave moment of the surface charge n( θ) distribution, where n(θ,z,t) = e j(ωt-kz) n(θ) . (D.1.4) These conditions determined the constants a m and Km giving the resulting E field inside the wire, E z(r,m) = (1/4) η m I Rdc (aβ') [ Jm(x) Jm+1(xa) - Jm(x) Jm-1(xa) ] a = radius ηm ≡ Nm N0 (D.2.33) E r(r,m) = (j/4) η m I Rdc (ak) [ Jm+1(x) Jm+1(xa) + Jm-1(x) Jm-1(xa) ] x = β 'r E θ(r,m) = (1/4) ηm I Rdc (ak) [ Jm+1(x) Jm+1(xa) - Jm-1(x) Jm-1(xa) ] x a = β'a where R dc = 1/(πa2σ) is the DC resistance of the wire per unit length, and I is the amplitude of the current in the wire. Everything is an implicit function of frequency ω. It was noted that, for |k/ β'| << 1, the fields Er and Eθ are much smaller than E z, and this is the case for f ~ 100 GHz or below (but not too small). An implication of the solution is that the E fields inside the wire for each partial wave are described by a single parameter N m which is the surface charge moment noted above. If the other transmission line conductor(s) were to change their position relative to the round wire and/or to vary their cross sectional shape, the only effect this would have w ould be to adjust the set of parameters N m, and the solutions would still be given by (D.2.33) quoted above. Although the set {N m} is infinite, it seems likely that for reasonable shapes of the other conductor(s), the lowest few N m partial waves would provide a good approximation to the E fields inside the wire. Since Ohm's Law is assumed to apply inside the wire, one then knows in detail the current densities J z, Jr and Jz. The magnetic field B inside the wire is then also known and was calculated above. The lowest moment is always N 0 = (k/2πωa) I from (D.2.31). Appendix D: Fields inside a Round Wire 319 As an example, the following five-conductor transm ission line might be expected to have a strong m = 2 quadrupole surface charge moment N 2, Fig D.4 A critical ingredient of our solution is the assumption that E θ(r=a,m) = 0 and that is the subject now addressed. We present two somewhat different arguments as to why E θ(r=a,m) = 0. It should be noted that King in his Transmission-Line Theory book always assumes that any straight transmission line conductor cross section has an equipotential surface (a ring, see for example middle p 14, top 15, 25 bottom). Due to the presence of small tran sverse vector potential components, E θ= 0 and "equipotential" for the scalar potential φ are not the same thing. (a) The Quasi-Static Argument In electrostati cs, we are used to metal surfaces being e quipotentials. For example, if we put a point charge q near a metal sphere, it induces a surface charge on th at sphere. The electric field lines land on the sphere exactly perpendicular to the surface. One argues that if there were even some tiny E field component tangential to the surface, the surface charges would adju st their position to cancel out that tangential field. Since the situation is static, any adju stment has already been made. Since E tan = 0, the sphere's surface is an equipotential surface. If we were to then slowly move the charge q around (perhaps it rotates in a circle around the sphere), the surface charge instantly adjusts at each new position of q, and those E field lines remain perpendicular to the surface, and E tan = 0. While the charges are adjusting position, there is admittedly some very tiny surface current driven by some tiny E tan , but if we move the charge slowly, we are "quasi-static" and the approximation E tan ≈ 0 is very good. One might compare the ti me constant of the moving sphere (T, the period of q's revolution around the sphere) to the time constant of the surface charge adjustment. For copper the time constant is roughly the mean el ectron collision time which is on the order of 10-14 sec. The conclusion here is that for frequencies << 1014 Hz, the quasi-static situation prevails and then E tan ≈ 0 is a very good approximation. This then is our first argument fo r why we claim the boundary condition E θ = 0 on the surface of the round wire in a transmission line operating at a typical frequency. We note from our solution E θ(r,m) that if we assume E θ(a,m) = 0 on the round wire surface, we will still have E θ(r,m) ≠ 0 inside the wire. This fact is consistent with our argument above since there are no free charges available to adjust themselves inside the wire. However: if E θ = 0 by this quasi-static argument, then we should expect that E z = 0 by the same argument, since E z is also a tangential field at the round wire surface, and since E z operates at the same frequency ω as Eθ. But we know that E z ≠ 0 because J z ≠ 0 just below the wire surface -- there is current Appendix D: Fields inside a Round Wire 320 flowing there -- and E z is continuous through the surface by (1.1.50). So the E field lines are not quite perpendicular to the round wire surface in the z dir ection. This is not too surprising since we expect everything to vary in the z direction as ej(ωt-kz) so we would expect the surface not to be an equipotential in this direction. But what happened to that quasi-sta tic argument we just applied to E θ ? What happened is that there is external field activity associated with the wave going down the line which forces E z ≠ 0. One might say the EM wave traveling down the line induces a J z in the round wire, with its associated E z ≠ 0. But then perhaps this same thing could somehow happen with E θ and then our quasi-static argument that E θ = 0 collapses. We think this could happen in fact, but only if the transmission line is driven by an apparatus which creates a "torsion wave" in the line. For ex ample, the apparatus could drive counter-rotating azimuthal currents onto the round wire surfaces of a twin-lead transmission line as suggested by this picture (which is not meant to imply that other field components vanish), Fig D.5 It seems from our work above that such a wave woul d satisfy Maxwell's equations and be a viable mode of the transmission line. In this case, E θ≠ 0 because the EM wave going down the line forces E θ ≠ 0, just as the normal wave forces E z ≠ 0. We have not investigated whether this type of tors ion wave is really viable. Whether or not it is, we assume in our transmission line discussion that this m ode is not activated and that therefore the quasi- static argument for E θ = 0 is valid at the round wire surface. (b) An Ansatz Argument We make an ansatz that E r,Eθ << Ez in our round wire E field solution, perhaps based on an expectation that most current in the wire will be longitudinal. We assume this is true, and see if this assumption is born out in a final solution of Ma xwell's equations. Given that E θ is then very small, we can make an approximation (another ansatz) that this field E θ is exactly zero on the surface of the round wire. This may not be exactly true, but again we assume it for our purposes and see where it leads. This is the nature of an "ansatz". When we make this assumption, the cross section of the transmission line may be regarded (Chapter 5) as a two dimensional potential theory problem -- basically a capacitor problem where one conductor has potential V/2 and the other -V/2, say (for a symmetr ic line, at some fixed value of z). In such a potential problem, one always assumes that the electrostatic potential φ is a constant on the surface of Appendix D: Fields inside a Round Wire 321 each conductor, and that is precisely what our ansatz says: E θ = -( ∇φ)θ = 0 , φ = constant in the θ direction. Now when we solve th e capacitor problem for potential φ, that gives E = - ∇φ in the dielectric between the conductors, and from that we may deduce E • n^ at the surface of one of the conductors. For the round wire with a cylindrical coordinate system whose axis is aligned with the wire center, that field is Er. Next, from this surface value of E r (which will be proportional to V) we may compute the surface charge density n( θ) on the round wire using (D.2.24) which says E r(r=a,θ) = (jω/σ) n(θ). For a "fat" twin lead transmission line for example we expect this to have a bulge in n( θ) on the side of the wire facing the other wire (m = 1, dipole), since that is what ha ppens in such a capacitor. In any event, given n( θ) we may compute the moments N m of the surface charge using (D.1.5b) and this then provides one "boundary condition" on our coefficients a m and Km which appear in all the field expressions we found above, E r(r=a,m) = (j ω/σ) Nm . (D.2.26) But recall that, in order to carry out this entire proces s just described, we had to start with the assumption that Eθ = 0 on the conductor cross section surface, so that we could have a capacitor problem in the first place. According to (D.1.3b), if E θ(r=a,θ) = 0, then E θ(r=a,m) = 0, so that in fact we must have E r(a,m) being zero in all partial waves m. Thus our assumed ansatz condition is Eθ( r = a , m ) = 0 (D.2.27) which is then a second boundary condition on a m and Km. Although (D.2.27) might not be exactly true, we know it is very close to being true. More importa ntly, we know that the above two conditions on a m and Km are consistent with each other, even though bot h boundary conditions might be slightly wrong. We then expect them to give good values for constants a m and Km. Using these "perhaps slightly wrong" boundary cond itions, we obtain the solutions shown in (D.2.33). It has already been noted above that for copper conductors and normal dielectrics, |k/ β'| << 1 up to at least 100 GHz. The condition |k/ β'| << 1 when applied to the (D.2.33) resu lts shows that in fact our ansatz that Er,Eθ << Ez is born out. There are three footnotes to the above discussion. First, we note that the second bo undary condition does not force E θ(r,m) = 0 for r < a inside the wire. In fact, there will be some small azimuthal "swirling" current inside the wire even if E θ(r=a,m) = 0, and this is just a result of Maxwell's e quations and their solutions above. Second, one might make the argument that the round wire surface is an equipote ntial since that is the way a line is driven at the source. For example, the center conductor of a coaxial cable plugs into a tiny driving cylinder (jack) in a BNC connector and this drives only the wire surface, and it does so in an azimuthally symmetric way so that one expects to have the wire surface be an equipotential at the driving point; this equipotential surface then mo ves down the line as the wave progresses. Third, we have the complication that we don't real ly have a purely electrostatic situation, and the potential is in fact related to E by equation (1.3.1) which says E = - ∇φ - ∂tA . The rescue here comes by claiming that roughly A ≈ Az^ so that the transverse components A r and A θ are very small. In this case, we then do get E ≈ -∇φ so that E θ = 0 is associated with constant φ on the wire surface. The argument for A ≈ Az^ is that A is driven by J, and J is mostly in the z^ direction, which in turn is related to our starting ansatz (see Appendix M). Appendix D: Fields inside a Round Wire 322 D.9 About the boundary condition E r(a,θ) = (jω /σ) n(θ ) . The "charge pumping boundary condition" appears in (D.2.23) and here we want to examine it more closely. Our concern is that the derivation of (D.2.23) ignores surface currents that we know exist on the surface of a transmission line conductor as the surface charge moves around in response to tangential E fields. The first issue then is to define and quantify the nature of these surface currents. (a) The notion of Debye Surface Currents We continue in the context of our classi cal treatmen t of the conductor surface. In Appendix E it is pointed out that the surface charge on a transmission line conduc tor exists in an incredibly thin surface layer we shall call the Debye layer for want of a better name. For copper the thickness λD of this layer is on the order of one atomic radius. In addition to the norma l conduction electrons, this thin layer contains extra free electrons that are piled up just below the surface (negative surface charge) or are depleted from this thin region (positive surface charge), as shown by th e red curve in Fig E.1. We want first so show : Fact 1: In a good conductor, the volume density of free electron carriers piled up at a surface (to make up the surface charge) is negligible compared to the volume density of conduction electrons. (D.9.1) Proof: From (E.7) the free charge density in the Debye layer (assume x is the inward surface normal direction) is given by ρ(x) = ρ(0) e-x/λD . The effective free surface charge n is then given by n = ∫0 ∞ dx ρ(x) = ρ(0) ∫0 ∞ dx e-x/λD = ρ(0)λD . The free electron density n e is then n e = ρ(0)/e = n / (e λD) . As a typical example, consider a parallel plate capacito r with close plate spacing s. The E field in the gap is E = V/s and the surface charge density from (1.1.47) is n = ε E = εV/s. For V = 10 volts and s = 1 mm we find n = ε 0V/s = 8.85 x 10-12 * 10 / 10-3 ≈ 101-12+1+3 = 10-7 Coul/m2 . Then the free electron density is ne = n / (eλD) ≈ 10-7 Cou/m2 / [ 1.6 x 10-19 Coul * 10-10m] ≈ 0.6 * 10 -7+19+10 ≈ 1022 electrons/m3 As noted in (N.1.2), in copper the conduction electron density (one electron per atom) is 10 29 /m3, QED. Corollary: The conductivity σ D inside the Debye layer is basically the same as σ outside that layer. ( D . 9 . 2 ) Appendix D: Fields inside a Round Wire 323 Proof: From (N.1.10) conductivity is σ = (nq2τ/m) where n is the electron density. The Fact above shows that this density is the same in the Debye layer as in the bulk conductor, so σD = σ . (We ignore the possibility that the collision time τ could differ in the Debye layer vs. in the bulk volume. ) QED Consider now this crude drawing which shows a tiny slice of width dx of a piece of a transmission line conductor cross section at its surface. The yellow Deby e surface charge layer is greatly exaggerated in thickness and is modeled as if it had a clean lowe r boundary. Recall from the comment below Fig E.2 that at 100 GHz one has δ ≈ 4000 λD so δ >> λD at all frequencies of transmission line interest. Fig D.6 The Debye layer holds the surface charge, and when th is surface charge moves, one has a Debye surface current. We now show : Fact 2: The total current in the Debye layer is negligible compared to that in the skin effect layer. ( D . 9 . 3 ) Proof : The field E z is parallel to the conductor surface, so we know from (1.1.41) that it is continuous through the boundary at the bottom of the Debye layer. Then the ratio of the currents in the two layers is, ID Iδ = JzD λd dx Jzδ δ dx = σDEzD λd σEzδ δ = σD σ Ez Ez λD δ ≈ 1 * 1 * λD δ = λD δ << 1 . QED (b) The role of Debye Surface Curr ents in the boundary conditi on Now referring to the Debye surface currents as K zD and K θD we reconsider the derivation of the charge pumping boundary condition of (D.2.24) where we had this figure, Appendix D: Fields inside a Round Wire 324 Fig D.2 If we include the Debye surface currents in the θ and z direction in our application of continuity, div J = - jωρ ⇔ -jω [∫V ρ dV] = ∫S J • dS , (D.2.22) the result is -jω n(θ,z) = - J r(r=a-ε, θ, z) + ∂zKz(θ,z) +(1/a) ∂θKθ(θ,z) (D.9.4) where we assume that the dielectric outside the round wire is vacuum with σ d = 0. The gaussian box selected here is that shown in red in Fig D.6. The bottom face lies belo w the Debye layer so J r(a-ε, θ, z) is the value of J r in the normal skin effect region close to the surface. The Debye surface currents may be written approximately as KzD = JzD λD = σDEzD λD = σ Ez(r=a,θ) λD // dim(K) = amp/m K θD = JθD λD = σDEθD λD = σ Eθ(r=a,θ) λD = 0 // (D.9.2) and (3.7.0) (D.9.5) where we use the Corollary above that σ D = σ. From (3.7.0) we have E θ = 0 at the surface so K θD = 0 and we have only the Debye current K zD to worry about. Recall that E θ(a,θ) = 0 is the second boundary condition (D.2.27) used in Section D.2 to evaluate the a m and Km coefficients, and that this condition is itself a topic of interest in Section D.8, and we assume it is valid. We then have, -jω n(θ,z) = - J r(a,θ,z) + ∂zKzD(θ,z) = - J r(a,θ,z) + ∂z [σ Ez(a,θ,z) λD] ( D . 9 . 6 ) Assuming everything has z dependence e j(ωt-kz) as in (D.1.4), we replace ∂z → -jk and then suppress the z arguments to get -jω n(θ) = - J r(a,θ) -jk [σ Ez(a,θ) λD] = - σEr(a,θ) -jk [σ Ez(a,θ) λD] = - σEr(a,θ)[ 1 - jkλ D Ez(a,θ) Er(a,θ) ] . (D.9.7) Appendix D: Fields inside a Round Wire 325 If we assume that the second term in (D.9.7) can be ignored, we get the charge pumping boundary condition E r(r=a-ε,θ) = (jω/σ) n(θ) (D.2.24) (D.9.8) which in return yields the E fields as stat ed in (D.2.33) where we see that roughly Ez(a,θ) Er(a,θ) ~ | β k | . (D.9.9) Thus, our self-consistent condition for ignoring the second term in (D.9.7) is k λD * Ez(a,θ) Er(a,θ) << 1 ⇔ k λD * | β k | << 1 ⇔ λD |β| << 1 ⇔ λD | ej3π/4 (2 /δ)| << 1 ⇔ (λ D/δ) << 1 ⇔ (δ/λd) >> 1 // ignore 2 But we know from above that ( δ/λd) >> 1 for any f < 100GHz, so for such f the second term in (D.9.7) can in fact be ignored. We have just proven: Fact 3: For f < 100 GHz, the Debye surface currents can be ignored in the derivation of the boundary condition E r(a-ε,θ) = (jω/σ) n(θ) . ( D . 9 . 1 0 ) (c) Where does surface charge n(θ ) come fro m? According to our traveling-wave ansatz (D.1.1), all E field related quantities move down a transmission line at v d as ej(ωt-kz). For a low-loss line and a vacuum dielectric, v d ≈ c, the speed of light. Therefore, n ( θ,z,t) = n(θ,0,0) ej(ωt-kz) k = ( ω/vd) . (D.9.11) One can ponder and then discard a list of hypotheses concerning where n( θ) "comes from" as it increases and decreases over time at some location z on one of the conductors. The first hypothesis might be that the individual electrons which make up n( θ) simply travel at v d in the z direction down the conductor surface, and n( θ) is not fed by any radial currents inside the conductor. In this case one would have K zD(θ) = vd n(θ). But we know this is not what happens. Apart from the massive energy required to achieve relativistic electron velocities, we know from Appendix N.1 that the electrons in the Debye layer in fact drift along at something like ~ 1 mm/sec, just as do the regular conduction electrons in the conductor bulk. The second hypothesis is a variation of the first, where we now allow that the Debye surface current works like any other conduction current, and when one electron moves "to the right" at some point z, a distant electron at z+L moves to the right at nearly the same time, all electrons in a long string moving to Appendix D: Fields inside a Round Wire 326 the right one position, giving the illusion that a particular electron moved very fast. This does in fact happen, and if it were all that happened, again we would have K zD(θ) = vd n(θ). Comment : Assume some skin depth δ ≤ a/10 so the bulk current is flowing in a sheath of thickness δ just under the conductor surface. The total sheath current is then roughly 2 πaδJz(a,θ). We can regard this current flow as due to an effective "full surface current" K z = Jz δ. Notice that this "surface current" is different from the "Debye surface current". Based on Fact 2 above, we certainly expect K z >> KzD . A third hypothesis is that somehow n( θ,z,t) is fed by azimuthal Debye surface currents, or some combination of these along with the z-directed K zD(θ). Our condition (3.7.0) that E θ = 0 puts a stop to the possibility of feeding by azimuthal Debye surface currents. What we have learned from Fact 3 is that none of the above hypotheses explains where n( θ) comes from. The analysis above shows that, although the surf ace motions of the Debye surface charges do create Debye surface currents, these currents are so small that they play no role in div J = -∂ tρ for the Gaussian box shown in red in Fig D.6. The charge n( θ) "comes from" inside the wire a nd is fed by the radial current density J r just below the surface according to (D.2.24), Jr(r=a-ε,θ) = jω n(θ) n( θ) = (1/jω) Jr(a-ε,θ) (D.9.12) Here is a suggestive picture, F i g D . 7 where the white boxes are little "radial charge pumps" delivering the required J r needed to feed the changing surface charge n( θ). Apart from the miniscule K zD , charges in n( θ) don't move in the z direction in this picture, they just appear to be doing that due to the choreographed radial pumping in and out at the wire surface. A wave front of the n( θ) wave travels at v d, and this is just a phase velocity. In an analogous situation, in a deep ocean wave the individual partic les of water travel in small ellipses and do not travel along with the wave, though there are small scal e longitudinal motions due to those ellipses. Comment : In our transmission line theory, the exterior problem in the dielectric is solved using the capacitor method, from which one learns n( θ). The boundary condition J r(r=a-ε,θ) = jω n(θ) couples this exterior information into the wire interior, allowi ng one to solve for the fields and currents inside. Appendix D: Fields inside a Round Wire 327 In Section 6.5 we show how div E = 0 inside the conductor (or div J = 0) forces a relationship between J r and Jz just below the conductor surface. When that rela tionship (6.5.18) is combined with the charge pumping boundary condition (D.9.8), one finds that Ez(a,θ) = (-jω/σ) (β/k) n(θ) (6.5.19) or J z(a,θ) = (-jω) (β/k) n(θ) . ( D . 9 . 1 3 ) This same result is obtained in a different manner as (6.5.13). The "full surface current" K z was defined in a Comment above as K z(θ) = δ Jz(a,θ). Thus, K z(θ) = δ Jz(a,θ) = [δ (-jω) (β/k)] n(θ) ( D . 9 . 1 4 ) But β k = ej3π/4 (2 /δ) ω/vd so [δ (-jω) (β/k)] = δ (-jω) ej3π/4 (2 /δ) ω/vd = -j ej3π/4 2 vd = -j (j-1)/ 2 * 2 vd = (1+j) v d and we end up with K z(θ) = (1+j) v d n(θ) Re(K z(θ)) = Im(K z(θ)) = vd n(θ) ( D . 9 . 1 5 ) Once again, this last equation gives the illusion that the surface charge density n( θ) moves "to the right" at speed v d to create the real or imaginary part of the full δ-thick surface current K z. This is the equation that replaces the incorrect equation K zD(θ) = vd n(θ) which assumes there is no radial charge pumping. Reader Exercise: Show using F = m a and F = qE (ignore magnetic fields) th at with a time-harmonic E field, a classical electron inside a transmission line conductor traverses a tiny elliptical path and thus never really goes anywhere. That path is traversed once per period T= 2π /ω. Mathematically, show that this amounts to proving that the three equations x = Acos( ωt-a) y = Bcos( ωt-b) z = Ccos( ωt-c) Appendix D: Fields inside a Round Wire 328 are parametric equations for an ellipse with some orientation in 3D space. As just noted above, this goes- nowhere aspect of the electron is similar to what happ ens with a droplet of water in an ocean wave. (Hint: first show that the first two equations describe an ellipse in the xy plane and that the semi-major axes in general are not A and B .) (d) Modifications for a Conducting Dielectric Ignorin g the Debye surface currents as per section (b) above, if the dielectric has some conductivity σd, the charge pumping boundary condition (D.2.23) becomes Jr(a-α,θ) - Jr(a+α,θ) = jω n(θ) or σ E r(a-α,θ) - σdEr(a+α,θ) = jω n(θ) // this is div J = -j ωρ where α > 0 is a tiny distance ( ε is already used for dielectric constant). Another boundary condition at the surface is provided by (1.1.47) which says ( n^ points into medium 1 which is the dielectric) [ε1En1 - ε2En2] = nfree // this is continuity of D n at the surface or [ε dEr1 - ε0Er2] = n(θ) or [ε d Er(a+α,θ) - ε0 Er(a-α,θ)] = n(θ) . A seeming third boundary c ondition is (1.1.48), ξ1En1 = ξ2En2 or (ε d + σd/jω) Erd = (ε0 + σ/jω) Er ≈ (σ/jω) Er or (ε d + σd/jω) Er(a+α,θ) ≈ (σ/jω) Er(a-α,θ) . There seem to be three boundary conditions at the round wire surface, σ E r(a-α,θ) - σdEr(a+α,θ) = jω n(θ) // modified cpbc from div J = -jωρ (D.9.16) εd Er(a+α,θ) - ε0 Er(a-α,θ) = n(θ) // div D = ρ (straddle) (D.9.17) (εd + σd/jω) Er(a+α,θ) ≈ (σ/jω) Er(a-α,θ) // ξ1En1 = ξ2En2 (D.9.18) but only two of these conditions are independe nt. For example, multiply (D.9.16) by (-1/j ω) to get (σd/jω)Er(a+α,θ) - (σ/jω) Er(a-α,θ) = - n(θ) . Adding this to (D.9.17) then gives Appendix D: Fields inside a Round Wire 329 (εd + σd/jω) Er(a+α,θ) - (σ/jω) Er(a-α,θ) = 0 which is in fact the same as (D.9.18). When we solve the "capacitor problem" as in Section 6.5 (a) to obtain n( θ) on the round conductor surface, we are using (D.9.17) with the assumption that E r(a+α,θ) >> Er(a-α,θ). Typically one just says that in a good conductor E r(a-α,θ) = 0 and then n( θ) = εd Er(a+α,θ). That is fine, but it is not clear what happens to (D.9.16) above. The first term is the product of a large quantity σ times a small quantity E r(a- α,θ) so can be the same size as the other terms in the equation. The resolution is provided by the discussion in S ection 1.5 (c) where we encountered the equation (1.5.17) n c(x,ω) = (ξ1/ε1) ns(x,ω) . (1.5.17) which in our current context (1 = dielectric) becomes n c(θ) = (ξd/εd) n(θ) . ( D . 9 . 1 9 ) In that discussion it is noted that n( θ) is the actual free surface charge density, whereas n c(θ) is a related "transport charge density" havi ng the same dimensions as n(θ ). If we multiply (D.9.16) and (D.9.17) by (ξd/εd), our (redundant) triplet of boundary conditions becomes, 1 σ (ξd/εd) Er(a-α,θ) - σd (ξd/εd) Er(a+α,θ) = jω nc(θ) 2 ξd Er(a+α,θ) - (ξd/εd) ε0 Er(a-α,θ) = nc(θ) 3 ξdEr(a+α,θ) ≈ ξ Er(a-α,θ) . ( D . 9 . 2 0 ) We now use the last of these three equations to eliminate E r(a+α,θ) in the first, which then becomes σ (ξ d/εd) Er(a-α,θ) - σd (ξ/εd) Er(a-α,θ) = jω nc(θ) or [ σ ξ d - σd ξ ]/εd * Er(a-α,θ) = jω nc(θ) or [ σ (ε d + σd/jω) - σd (ε0 + σ/jω) ]/εd * Er(a-α,θ) = jω nc(θ) or [ (σ ε d - σdε0)]/εd * Er(a-α,θ) = jω nc(θ) // two large terms cancelled or [ σ - σ d(ε0/εd) ] Er(a-α,θ) = jω nc(θ) . Assume now that ε 0 (conductor) and εd (dielectric) are the same order of magnitude, and assume that, even though the dielectric conducts, one still has σ >> σd . The last equation then reads E r(a-α,θ) = (jω /σ) nc(θ) = (jω /σ) (ξd/εd) n(θ) ( D . 9 . 2 1 ) Er(a-α,m) = (jω/σ) (ξd/εd) Nm . // partial waves (D.9.22) Appendix D: Fields inside a Round Wire 330 These are the the "modified" charge pumping boundary conditions which replace (D.2.24) and (D.2.25) for a mildly conducting dielectric, E r(r=a,θ) = (jω/σ) n(θ) . (D.2.24) Er(r=a,m) = (j ω/σ) Nm . (D.2.25) How then does σ d ≠ 0 alter the E field results summarized in box (D.2.33)? The rule is this: N m → (ξd/εd) Nm everywhere . For example, for the total current we have I = 2 πaN 0ω/k → I' = 2 πaN0(ξd/εd)ω/k' . The current appears to grow larger due to the new factor (ξ d/εd), but one must realize that the wavenumber k also changes to k' when the dielectri c conduction is turned on. Using the model of Chapter 4, we have k = -j (R+jωL)(jωC) → k' = -j (R+jωL)(G+jωC) . Therefore the new current may be written in terms of the old current, I' = I ( ξd/εd) (k/k') = I ( ξd/εd) jωC G+jωC . For ω >> G/C, I' is larger than I by the full factor ( ξd/εd). If we now denote by I the actual current in the transmi ssion line (I for G = 0, or I' for G>0) , then the E field solutions as expressed in (D.2.33) do not change form , and we may write Second summary of the E field solutions : R dc = 1 σπa2 β'2 = β2 - k2 (D.2.33) conducting dielectric Ez(r,m) = (1/4) η m I Rdc (aβ') fm f m = [ Jm(x) Jm+1(xa) - Jm(x) Jm-1(xa) ] x = β 'r Er(r,m) = (j/4) η m I Rdc (ak) gm g m = [ Jm+1(x) Jm+1(xa) + Jm-1(x) Jm-1(xa) ] x a = β'a Eθ(r,m) = (1/4) ηm I Rdc (ak) hm h m = [ Jm+1(x) Jm+1(xa) - Jm-1(x) Jm-1(xa) ] Appendix D: Fields inside a Round Wire 331 D.10 High frequency limit of the round wire E fields The box (D.2 .33) above displays the round wire E i field components in terms of functions f m., gm and hm. Here we study first the symmetry under m ↔-m of these functions, and then we evaluate them at high frequency. (a) Symmetry of f m, gm and hm and expansions for E i(r,θ) From NIST (10.4.1) we know that for integer m, J -m(x) = (-1)mJm( x ) . ( D . 1 0 . 1 ) For f m we find that fm = [ Jm(x) Jm+1(xa) - Jm(x) Jm-1(xa) ] f-m = [ J-m(x) J-m+1(xa) - J-m(x) J-m-1(xa) ] = - [ Jm(x) Jm-1(xa) - Jm(x) Jm+1(xa) ] = fm Coefficients g m and hm have the same symmetry, gm = [ Jm+1(x) Jm+1(xa) + Jm-1(x) Jm-1(xa) ] g -m = [ J-m+1(x) J-m+1(xa) + J-m-1(x) J-m-1(xa) ] = [ Jm-1(x) Jm-1(xa) + Jm+1(x) Jm+1(xa) ] = gm hm = [ Jm+1(x) Jm+1(xa) - Jm-1(x) Jm-1(xa) ] h-m = [ J-m+1(x) J-m+1(xa) - J-m-1(x) J-m-1(xa) ] = [ Jm-1(x) Jm-1(xa) - Jm+1(x) Jm+1(xa) ] = - [ Jm+1(x) Jm+1(xa) - Jm-1(x) Jm-1(xa) ] = hm Thus we have shown that f-m = fm g-m = gm h-m = hm . ( D . 1 0 . 2 ) If the surface charge n( θ) happens to be even in θ, we know from (D.1.7) that η m = Nm/N0 = η-m. In this case, the E field components in (r, θ) can be written as in (D.1.7), Appendix D: Fields inside a Round Wire 332 Ei(r,θ) = Ei(r,m=0) + 2 ∑ m = 1∞ Ei(r,m) cos(m θ) (D.10.3) Then for even n( θ) the E fields are Ez(r,θ) = (1/4) I R dc (aβ') [ f0 + 2 Σm=1∞ fm ηm cos(mθ) ] E r(r,θ) = (j/4) I R dc (ak) [ g0 + 2 Σm=1∞ gm ηm cos(mθ) ] Eθ(r,θ) = (1/4) I R dc (ak) [ h0 + 2 Σm=1∞ hm ηm cos(mθ) ] . (D.10.4a) For general n( θ) where ηm and η-m are no longer equal we have instead E z(r,θ) = (1/4) I R dc (aβ') [Σm=-∞∞ fm ηm ejmθ ] E r(r,θ) = (j/4) I R dc (ak) [ Σm=-∞∞ gm ηm ejmθ ] E θ(r,θ) = (1/4) I R dc (ak) [ Σm=-∞∞ hm ηm ejmθ ] . (D.10.4b) (b) High frequency evaluation of f m, gm and hm and the E fields For large ω we can use the following expressions for β' and k : β2 = -jωμσ (1.5.1d) for good conductor k ≈ ωLC - j (RC + GL) 2LC ≈ ω/vd = βd (Q.2) β'2 = β2 - k2 ≈ -jωμσ - ω2LC . Although the second term appears to win out here for large ω, for frequencies of interest to us the first term is always much larger due to the large size of σ (see discussion below (D.2.2)) . Therefore β' ≈ β ≈ -jωμσ . At high ω this β ' parameter is very large, so x a = β'a will also be large and we need then to find the high ω limits of our functions f m, gm and hm . From NIST 10.17.2, keeping a few leading terms in each inverse power expansion, we have this rather complicated large x behavior for J m(x), Jm(x) = (2/πx)1/2 { cos(w) [a 0(m) - a2(m)/x2 + O(1/x4)] - sin(w) [a 1(m)/x + O(1/x3)] ] } w = x - m π/2 -π/4 => e -jw = e-j(x-mπ/2-π/4) = e-jx ejπm/2 ejπ/4 Appendix D: Fields inside a Round Wire 333 a0(m) = 1 a 1(m) = 4m2-1 8 ≡ cm a 2(m) = (4m2-1)(4m2-9) 128 ≡ dm . (D.10.5) The expansion is in fact valid for all real and comp lex values of the parameter m, but we shall only use the expansion for integer m. Using abbreviations c m and dm one gets, Jm(x) = (2/ πx)1/2[ cos(w) (1-d m/x2) - sin(w) (c m/x ) ] . (D.10.6) Recall that inside the round wire, δ ≡ 2/ωμσ = skin depth // ωμσ = 2/δ2 (2.2.20) β = ej3π/4 (2 /δ) = (j-1)/δ (2.2.21) so x = βr = e j3π/4 (2 /δ) r = (j-1) (r/ δ) xa = βa = ej3π/4 (2 /δ) a = (j-1) (a/δ ) . ( D . 1 0 . 7 ) Since x has a large positive imaginary part for small δ, so does w. Then cos(w) = [ ejw + e-jw]/2 ≈ (1/2) e-jw sin(w) = [ ejw - e-jw]/2j ≈ -(1/2j) e-jw = (j/2)e-jw . (D.10.8) The large-x expansion above then becomes J m(x) = (2/π x)1/2 (1/2) [e-jw (1- dm /x2) - j e-jw (cm /x) ] = ( 1 / 2 πx)1/2 e-jw [ 1 -j cm (1/x) - d m (1/x2) + ... ] = ( 1 / 2 πx)1/2 e-jx ejπm/2 ejπ/4 [ 1 -j cm (1/x) - d m (1/x2) + ... ] . (D.10.9) It is not hard to show that this agrees with (2.3.5) through order 1/x. Notice from (D.10.7) that e -jx = e-j(j-1)(r/ δ) = e(1+j)(r/ δ) giving a convenient hybrid form Jm(x) = (1/2 πx)1/2 e(1+j)(r/ δ) (j)m ejπ/4 [ 1 -j cm (1/x) - d m (1/x2) + ... ] . (D.10.10) From (D.10.10) we see by inspection that, through O(1/x), J m(x) Jn(xa) = (j)m-n a r e(1+j)(r-a)/ δ1 -j cm (1/x)- d m (1/x2) + ... 1 -j cn (1/xa)- dn (1/xa2) + ... ≈ (j)m-n a r e(1+j)(r-a)/ δ [ 1 - jcm/x + jcn/xa ] (D.10.11) Appendix D: Fields inside a Round Wire 334 and Jm(x) Jm(xa) = a r e(1+j)(r-a)/ δ // independent of m (D.10.12) Therefore for large ω, g m = [ Jm+1(x) Jm+1(xa) + Jm-1(x) Jm-1(xa) ] = 2 a r e(1+j)(r-a)/ δ hm = [ Jm+1(x) Jm+1(xa) - Jm-1(x) Jm-1(xa) ] = 0 fm = [ Jm(x) Jm+1(xa) - Jm(x) Jm-1(xa) ] = (j)-1 a r e(1+j)(r-a)/ δ [ 1 - jcm/x + jcm+1/xa ] - (j)+1 a r e(1+j)(r-a)/ δ [ 1 - jcm/x + jcm-1/xa ] = - j a r e(1+j)(r-a)/ δ [ 2 - 2jc m(1/x) + j(c m+1+cm-1) (1/xa) ≈ - 2j a r e(1+j)(r-a)/ δ The results are then Ez(r,m) = (1/4) η m I Rdc (aβ) fm f m = -2j a r e(1+j)(r-a)/ δ x = β 'r Er(r,m) = (j/4) η m I Rdc (aβd) gm g m = 2 a r e(1+j)(r-a)/ δ x a = β'a Eθ(r,m) = (1/4) ηm I Rdc (aβd) hm h m = 0 or Large ω limits of the E field solutions : R dc = 1 σπa2 (D.10.13) Ez(r,m) = -(j/2) η m I Rdc (aβ) a r e(1+j)(r-a)/ δ x = β r Er(r,m) = (j/2) ηm I Rdc (aβd) a r e(1+j)(r-a)/ δ x a= βa Eθ(r,m) = 0 As observed earlier, the longitudinal current J z is much larger than the radial current J r by factor | β/βd|. Appendix D: Fields inside a Round Wire 335 Notice the standard skin effect behavior both in amplitude and phase for both field components. We saw this earlier in several places: E(x,ω) = E(0,ω ) e -x/δ e-jx/δ . x → (a-r) 1D example (2.1.8) |Ez(r)| |Ez(a)| = a r e(r-a)/δ r/δ > 3/ 2 = 2.1 . (2.3.7) The θ-space fields from (D.10.4b) are then, Ez(r,θ) = -(j/2) I R dc (aβ) { a r e(1+j)(r-a)/ δ } [ Σ m = -∞∞ ηm ejmθ ] Er(r,θ) = (j/2) I R dc (ak) { a r e(1+j)(r-a)/ δ } [Σm = -∞∞ ηm ejmθ ] Eθ(r,θ) = 0 ηm = Nm/N0 . (D.10.14) But the [...] expansions shown here are just n( θ)/N0 from (D.1.5a) where recall that (for G=0), N0 = (k/2πωa ) I . (D.2.31) Since Rdc = (πa2)/σ we find, (1/4) I R dc(ak) 2 /N 0 = (1/2) I R dc (ak) 2πωa/(kI) = R dc πωa2 = (ω/σ) so then for large ω (D.10.14) becomes, Ez(r,θ) = - (jω/σ) a r e(1+j)(r-a)/ δ n(θ) (β/k) Er(r,θ) = (jω/σ) a r e(1+j)(r-a)/ δ n(θ) Eθ(r,θ) = 0 ( D . 1 0 . 1 5 ) so both E z and Er track with n( θ). From (D.10.14) the surface impedance is then Zs(θ) ≡ Ez(r,θ) / I = -(j/2) R dc (aβ) n(θ)/N0 = -(j/2) R dc (aβ) n(θ) <n(θ)> . Since β = ej3π/4 (2 /δ), this can be written Zs(θ) = 1 σ(2πa)δ (1+j) n(θ) <n(θ)> . ( D . 1 0 . 1 6 ) Appendix D: Fields inside a Round Wire 336 and Zs(θ) is seen to track n( θ). Averaging over the round wire surface (as done in (4.11.9) ) then gives < Zs(θ)> = 1 σ(2πa)δ ( 1 + j ) ( D . 1 0 . 1 7 ) which is the same as Z s appearing in (2.4.16). For G≠0, one should add a factor ( ξ d/εd) to (D.10.15) through (D.10.17). Observations on the E fields for large ω In the extreme skin effect (small δ, large ω) regime: 1. There is no azimuthal field E θ inside or on the surface of the round wire. 2. Both E z and Er exhibit the standard skin effect form for amplitude and phase 3. At least for low loss situations, the ratio E z(r,θ)/Er(r,θ) = - (β'/k) = - (β'/βd) is very large in magnitude and is constant in r and θ. 4. Both E z and Er track the surface charge density n( θ) for azimuthal dependence 5. If n(θ) ≠ constant, then J z = σEz ≠ constant in θ and the longitudinal current density is asymmetric across the round wire cross section, which is known as th e proximity effect. This implies that the surface impedance Z s is a function of θ. In Section 2.4 the surface impedance was a constant since only the m=0 partial wave was involved. Appendix D: Fields inside a Round Wire 337 D.11 Low frequency limit of the round wire E fields (a) A High Level Review of Appendix D and its Accuracy As presented above, the general approach of Appendix D was to solve for the E and B fields inside a round wire assuming the ansatz traveling wave form E(r,θ,z,t) = e j(ωt-kz) E(r,θ) (D.1.1) where k is an arbitrary complex parameter. For any k , we found the following E field solution, where k dependence is now show n more explicitly: Second summary of the E field solutions : R dc = 1 σπa2 β'(k)2 = β2 - k2 (D.2.33) Ez(r,m) = (1/4) η m I Rdc [aβ'(k)] fm(k) f m(k) = [ Jm(x) Jm+1(xa) - Jm(x) Jm-1(xa) ] x = β '(k)r Er(r,m) = (j/4) η m I Rdc (ak) gm(k) g m(k) = [ Jm+1(x) Jm+1(xa) + Jm-1(x) Jm-1(xa) ] x a = β'(k)a Eθ(r,m) = (1/4) ηm I Rdc (ak) hm(k) h m(k) = [ Jm+1(x) Jm+1(xa) - Jm-1(x) Jm-1(xa) ] These fields exactly solve Maxwell's equations and the two boundary conditions (D.2.26) and (D.2.27), and from these E fields we computed the corresponding B fields. The coefficients ηm are the moments of the surface charge distribution n( θ) on the round wire surface. In principle, any linear combination of these solutions for different k values (including a cont inuous superposition) is also a possible solution. However, when this round wire is part of a tran smission line, one must also take into consideration the field solution outside the round wire -- the solution within the transmission line dielectric region. This is the so-called exterior solution, whereas our round wi re analysis provided an interior solution. The idea is that the exterior solution provid es the correct value of parameter k to use for the interior solution. The solutions must have the same k value due to the boundary between interior and exterior. Whereas Appendix D found the interior solution for the E field using the Helmholtz equation, Chapters 3 and 4 obtained the exterior solution in terms of the potentials φ and A z using the King gauge condition. This analysis was not valid at low frequencies for a variety of reasons noted in those chapters, perhaps the most dramatic of which is shown in Fig 3.6.(b). This drawing illustrates how the round wires of a twin-lead transmission line are clearly not surfaces of constant A z potential at very low frequency, whereas the theory assumes that they are. The main results of Chapter 4 were the first and second order "transmission line equations" (4.11.14b) and (4.1 1.15) involving i(z) and V(z). The second order equations are (damped, ω domain) wave equations which directly imply an e-jkz dependence on z. Through the boundary between the interior and exterior solutions, this implies a similar e-jkz form for the interior solutions, which form is the ansatz of Appendix D. However, at low frequencies these wave equations are no longer valid, there are "correction terms", and thus the e-jkz ansatz (D.1.1) of Appendix D is no longer valid. Therefore, we cannot expect low frequency predictions of Appendix D concerning interior fields to be accurate. Appendix D: Fields inside a Round Wire 338 Meanwhile, on a separate track altogether, Appendix K describes the so-called "network model" of the exterior solution [ at least i(z) and V(z) ] for a transmission line, using lumped R,G,L,C components. In this model, the same transmission line equations ob tained in Chapter 4 are found to be true, justifying the network model. However, in the network model, these transmission line equations are valid all the way down to DC ( ω=0) whereas we have just shown that the "physics model" does not support this conclusion. Nevertheless, we can use the network mode l's low frequency range as an approximation to the true exterior solution at low frequency. In other words, we can pretend that the transmission line equations are valid all the way down to DC. In so doi ng, we should not be surprised to find results which are inaccurate. Note that the network model says nothing about interior field solutions. Above low frequencies both the physics and network models provide the same value of k to be used in the round wire interior solution. That value is k = -j zy = -j (R+jωL)(G+jωC) . Since k is a function of ω (explicitly and also through ω dependence of the parameters), th e transmission line has "dispersion" and a group velocity v g = ∂ω/∂k different from the phase velocity v φ = ω/k. Appendix Q obtains expressions for k( ω) appropriate for both high and low frequenci es as limits of this rather complicated function. We then use these limits, knowing that th ey can give inaccurate results, in our low frequency analysis below. Appendix R makes use of the k( ω) function in a case study of a certain Belden cable. (b) Low frequency values for β' For low ω and G > 0, we use these expressions for β and k, β2 = -jωμσ (1.5.1d) for good conductor k ≈ (ω/2) RC+GL RG - j RG . (Q.3) so that at low ω β2 ≈ 0 k ≈ - jRG => k2 = - RG β'2 = β2 - k2 = -k2 ≈ R G . ( D . 1 1 . 1 ) For any reasonable transmission line RG will very small so the Bessel argument x a = β'a << 1. Low Frequency Example : Belden 8281 coaxial cable has radius a = 394 μ, K = 3.7 and R = 36.1 ohms/km ≈ .04 ohm/m. At a worst case 100 GHz it has a significant σd = σeff = 2.6 x 10-3 mho/m from (3.3.6) which we will conservatively assume applies also at low frequency (in reality σd is much less). From (4.11.34) G = 4 πσd/K = 4π 2.6 x 10-3/ K = .009 mho/m. Then RG = .04*.009 = .00036 and β' = RG ≈ .02 m-1. Finally, the maximum Bessel function argument in (D.2.33) is x a = β'a = .02 * 394e-6 ≈ 10-5. Since β'a is very small, we shall need to evaluate f m, gm and hm for small β '. Appendix D: Fields inside a Round Wire 339 For low ω and G= 0, we use instead these expressions β2 = -jωμσ (1.5.1d) for good conductor k ≈ RC/2 ω1/2 (1-j) = RC ω1/2 e-jπ/4 (Q.4) (highly damped) k2 = RCω(-j) = -jωRC β' 2 = β2 - k2 ≈ -jωμσ +jωRC = -jω (μσ - RC) (D.11.2) Again β' is very small at low frequency. (c) Low frequency evaluation of f m, gm and hm In the following we consider only m ≥ 0 since we know from (D.10.2) that f -m = fm , g-m = gm, h-m = hm. The small x limit for J m(x) is given by NIST 10.7.3, J n(x) = (x/2)n / n! . for n = 0,1,2,..... (D.11.3) Since J m-1 appears in our coefficient expressions and since m = 0 is encountered, we have to deal with m = 0 as a special case since the above li mit is not valid for n = -1. To this end we use NIST 10.2.2 which is valid for integer n, J-n(x) = (-1)nJn(x) ≈ (-1)n (x/2)n / n! (D.11.4) so that J -1(x) = - J1(x) ≈ - (x/2). Our small-x forms of interest are then J n(x) = (x/2)n / n! for n = 0,1,2,..... J-1(x) = - (x/2) for n = -1 . (D.11.5) We now examine the small x limits of f m, gm, and hm . First f m for m > 0, and then for m = 0: fm = [ Jm(x) Jm+1(xa) - Jm(x) Jm-1(xa) ] = [(x/2)m/m! (xa/2)m+1/(m+1)! - (x/2)m/m! (xa/2)m-1/(m-1)! ] = [ (m+1) (x/x a)m (2/xa) - (1/m) (x/x a)m(xa/2) ] = (x/x a)m [ (m+1) (2/x a) - (1/m) (x a/2) ] ≈ (x/x a)m (m+1) (2/x a) // as x a→ 0 f0 = [ J0(x) J1(xa) - J0(x) J-1(xa) ] = [ J0(x) J1(xa) + J0(x) J1(xa) ] = 2 J0(x) J1(xa) = 2 1 (xa/2) = 4/xa Appendix D: Fields inside a Round Wire 340 First gm for m > 0, and then for m = 0: gm = [ Jm+1(x) Jm+1(xa) + Jm-1(x) Jm-1(xa) ] = [(x/2)m+1/(m+1)! (xa/2)m+1/(m+1)! + (x/2)m-1/(m-1)! (xa/2)m-1/(m-1)! ] = ( x / x a)m+1 + (x/xa)m-1 g0 = [ J1(x) J1(xa) + J-1(x) J-1(xa) ] = [ J1(x) J1(xa) + J1(x) J1(xa) ] = 2 J1(x) J1(xa) = 2 (x/x a) Results for h m are then obvious since there is only a sign change between the terms in g m, h m = (x/xa)m+1 - (x/xa)m-1 h 0 = 0 The results are then, f m = (r/a)m (m+1) (2/ β'a) f 0 = 4/(aβ') gm = (r/a)m+1 + (r/a)m-1 g 0 = 2 (r/a) hm = (r/a)m+1 - (r/a)m-1 h 0 = 0 for m ≥ 0. Allowing for all integer values of m, using the symmetries (D.10.2) we can write, fm = (r/a)|m| (|m|+1) (2/β 'a) f 0 = 4/(aβ') gm = (r/a)|m|+1 + (r/a)|m|-1 g 0 = 2 (r/a) hm = (r/a)|m|+1 - (r/a)|m|-1 h 0 = 0 . (D.11.6) (d) Low frequency E fields The fields in (D.2.33) quoted above then become, Ez(r,m) = (1/4) η m I Rdc [aβ'] fm f m = (r/a)|m| (|m|+1) (2/β 'a) f 0 = 4/(aβ') Er(r,m) = (j/4) η m I Rdc (ak) gm g m = (r/a)|m|+1 + (r/a)|m|-1 g 0 = 2 (r/a) Eθ(r,m) = (1/4) ηm I Rdc (ak) hm h m = (r/a)|m|+1 - (r/a)|m|-1 h 0 = 0 or Ez(r,m) = (1/2) η m I Rdc (r/a)|m| ( | m | + 1 ) Er(r,m) = (j/4) η m I Rdc (ak) [(r/a)|m|+1 + (r/a)|m|-1] (D.11.7) Eθ(r,m) = (1/4) ηm I Rdc (ak) [(r/a)|m|+1 - (r/a)|m|-1] Ez(r,0) = I R dc Er(r,0) = (j/2) I R dc (ak) (r/a) Eθ( r , 0 ) = 0 / / l o w ω E fields Appendix D: Fields inside a Round Wire 341 If n(θ) is real and even in θ, we know from (D.10.4a) that, Ez(r,θ) = (1/4) I R dc (aβ') [ f0 + 2 Σm=1∞ fm ηm cos(mθ) ] Er(r,θ) = (j/4) I R dc (ak) [ g0 + 2 Σm=1∞ gm ηm cos(mθ) ] Eθ(r,θ) = (1/4) I R dc (ak) [ h0 + 2 Σm=1∞ hm ηm cos(mθ) ] . (D.10.4a) Inserting the expressions (D.11.6) then gives E z(r,θ) = I Rdc { 1 + Σm=1∞ (r/a)m (m+1) ηm cos(mθ) } Er(r,θ) = (j/2) I R dc (ak) {(r/a) + Σm=1∞ [(r/a)m+1 + (r/a)m-1] ηm cos(mθ) } Eθ(r,θ) = (1/2) I R dc (ak) { 0 + Σm=1∞ [(r/a)m+1 - (r/a)m-1] ηm cos(mθ) } . (D.11.8) As for surface impedance, from (D.11.8) we find that, for low ω, Zs(θ) ≡ Ez(a,θ)/I = Rdc { 1 + Σm=1∞ (m+1) ηm cos(mθ) } (D.11.9) <Zs(θ)> = Rd c (D.11.10) and this last result certainly seems reasonable. Notice that, E z(r,θ) Ez(r,0) = 1 + Σm=1∞ (r/a)m (m+1) ηm cos(mθ) . (D.11.11) An anomaly. This last result is supposedly valid for very low ω, and we see that E z(r,θ) and the above ratio are independent of ω since k does not appear anywhere. We would expect that in the limit ω→0 the above ratio should be exactly 1, at least for a non- conducting dielectric. This is so because we expect there to be no eddy currents at ω = 0 and these are the cause of J z non-uniformity as discussed in Appendix P. We attribute this anomalous result to the inaccuracy of the model at low ω, as outlined in section (a) above. It happens that in the limit ω→ 0 we also have I → 0 when G = 0 (since then Z 0 → ∞), but that is no justification for the anomalous result. We ignore this anomaly and proceed with our task of finding the low frequency E fields in more detail. Recall from (4.11.16) that Z 0 = R + jωL G + jωC 1/Z 0 = G + jωC R + jωL . (4.11.16) For low ω we find from these expressions for 1/Z 0 and from (D.11.1,2) that, I = V/Z 0 ≈ V G/R k ≈ -j RG G > 0 (D.11.12) I = V/Z 0 ≈ V jωC/R k ≈ RC ω1/2 e-jπ/4 G = 0 (D.11.13) Inserting (D.11.12) into (D.11.7) gives, for G > 0, Appendix D: Fields inside a Round Wire 342 Small ω limit of the E field solutions, G>0 : R dc = 1 σπa2 (D.11.14) Ez(r,m) = (1/2) η m V Rdc G/R (r/a)|m| (|m|+1) Z 0 = R/G Er(r,m) = a (1/4) η m V Rdc G [(r/a)|m|+1 + (r/a)|m|-1] Eθ(r,m) = -ja (1/4) ηm V Rdc G [(r/a)|m|+1 - (r/a)|m|-1] k = -j RG Ez(r,0) = V R dcG/R Er(r,0) = a (1/2) V R dc G (r/a) Eθ(r,0) = 0 The fields are all finite and there are non-zero expressions for E r and Eθ which account for the expected non-uniform flow of current into the dielectric through the conductor boundaries. We expect that E r(r,θ) just outside the conductor boundary is a strong function of θ for closely spaced conductors (the capacitor problem), and thus so is J r(r,θ). But Jr(a,θ) is continuous through the boundary at ω = 0, so we expect to see a strong dependence of J r(a,θ) on θ inside the round wire, as indicated by E r(r,m) in (D.11.14). Reader Exercise: Does (D.11.14) give the correct solution to the implied magnetostatics problem, or are there anomalies like the one noted above? Notice that Z 0 is certainly correct based on the reader exercise given in Appendix K (c). The "wave" decays in z according to e-jkz = exp(- RG z) which also seems reasonable. Next, inserting (D.11.13) into (D.11.7) gives, for G = 0, this limiting form : Small ω limit of the E field solutions, G=0 : R dc = 1 σπa2 (D.11.15) Ez(r,m) = (1/2) η m V ω1/2jC/R Rdc (r/a)|m| (|m|+1) Er(r,m) = a (j/4) η m V ω C Rdc [(r/a)|m|+1 + (r/a)|m|-1] Eθ(r,m) = a (1/4) ηm V ω C Rdc [(r/a)|m|+1 - (r/a)|m|-1] Ez(r,0) = V ω1/2jC/R Rdc Er(r,0) = a (j/2) V ω C Rdc (r/a) Eθ(r,0) = 0 As ω → 0, all fields vanish, corresponding to the fact that Z 0 = R/jω C → ∞ so I = V/Z 0 → 0. In this situation the network model is just an infinite ladder of series resistors with no conductance cross pieces. Fig D.8 Appendix E: How Thick is Surface Charge on a Metal? 343 Appendix E: How Thick is Surface Charge on a Metal Conductor? It is often said that surface charges exist only very clo se to the surface of a conductor. In this section, we will show how extremely true this statement is. Here is a crude sketch of what we expect surface charge distributions might look like at the plates of a capacitor. Fig E.1 The red plot is charge density ρ, and the black plot is the electric field magnitude. The charge density is exactly ρ = 0 in the dielectric region between the two plat es simply because there are no available charge carriers as there are in a metal (the electron cloud), see Section 3.1. Barring a huge E field or very high temperatures, electrons cannot just "jump off" the metal surface into the dielectric region because of an energy cost to do so called the work function. The figure suggests that the charge distributi on might have an exponential decay going into each metal surface, with some characteristi c distance which we seek to find. The reader might wonder: is it the skin depth δ? The answer to that question is: most definitely not! We are used to using Ohm's law J = σE in various forms. Application of this law in the regions of charge density in the above figure leads to a contradic tion. In the DC static case, nothing moves, so there can be no J, but there is clearly some E, so how can J = σE ? The reason is that Ohm's law only applies in a neutral medium. When there is a net charge density, the corrected Ohm's law is this: J = σE - D grad ρ . dim(D) = m 2/sec (E.1) The grad term, associated with Fick's Law, represents a flux of charged particles (a current) created by a gradient of the charge density. The charge flows (diffuses) from a region of high density to one of lower density, hence the minus sign, just as heat flows fr om a region of higher temperature to one of lower temperature. In a static situation with no current, the second term balances th e first term in a surface charge region, σE = D grad ρ . (E.2) As electrons pile up on the boundary, they resist furthe r pileup by their higher density. Basically this is a diffusion effect, and D is a diffusion coefficient. There is another more familiar equation which relates E and ρ, namely (1.1.3) + (1.1.6), div E = ρ/ε . ( E . 3 ) Appendix E: How Thick is Surface Charge on a Metal? 344 Inside a metal conductor the dielectric constant ε requires some careful study, but here we shall just set it to ε0 as if there were nothing in the electron cloud of the metal that could be polarized. Taking the divergence of (E.2) and using (E.3) we get this result ∇2ρ = (σ /Dε0) ρ . (E.4) The inverse combination of symbols in (E.4) is the square of something called the Debye length, λD2 = (Dε0/σ) (E.5) which is associated with charge screening in plasmas (such as the electrons in a metal). Thus, (E.4) may be written, ∇ 2ρ = (1/λD2) ρ . (E.6) In our one-dimensional problem of the above figure, the solution of this equation is ρ(x) = ρ(0) e-x/λD ( E . 7 ) where x is a coordinate going into the surface. This says that the thickness of the charge surface layer inside the metal is basically λD. If the electron cloud inside the metal is treated as a classical gas of particles of mass m, charge q, temperature T, and density n, one gets formulas for the various coefficients. Here are some expressions: J = nqv v = average drift velocity // (N.1.1) τ = mean lifetime between collisions // below (N.1.2) μ = (v/E) = (q/m) τ = mobility // (N.1.7) D = kT( μ/q) = kT(τ/m) = diffusion coefficient // " Einstein relation" σ = (nq 2τ/m) = conductivity // (N.1.9) λD = ε0 kT/nq2 = Debye length // (E.5) and last 2 equ. above (E.8) This set of equations represents a classical model for the free charge in a metal. One major and one minor adjust ment is needed (see Kittel p 278-280) when quantum theory is applied because electrons are fermions. This means that they cannot all park in the same state, so they "pile up" in higher and higher states in something known as the Fermi sphere. Only electrons at the surface of this sphere can do anything useful. Due to the pileup, the temperature of the active electrons is very much higher than one might think using classical physics. One finds this temperature by setting kT = EF where this latter is the Fermi energy, E F = (h2/ 8π2m) (3π2n)2/3 = kTF . ( E . 9 ) Appendix E: How Thick is Surface Charge on a Metal? 345 The appearance of the Plank constant h is the clue that this is a quantum result. This was the major quantum adjustment. The minor one is that T in th e Debye formula gets replaced by (2/3)T. Thus, λD = ε0 k(2TF/3)/nq2 = Debye length (quantum correct) . (E.10) We shall now run some numbers. Here are the basics, n = 8.45 x 10 28 electrons/ m3 for Copper k = 1.38 x 10 -23 = Boltzmann constant m = 9.1 x 10-31 kg = electron mass h = 6.63 x 10-34 J sec = Planck constant Plugging these into (E.9) gives the following effective electron temperature so TF = 81,702 ° K = pretty hot . (E.11) We can now compute the Debye length, using (E.10) : ε 0 = 8.85 x 10-12 F/m q = 1.60 x 10 -19 C so λD = 5.55 x 10-11 m = 0.55 A (Angstroms) // = 55 pm (E.12) and this result for λD appears on page 280 of Kittel. The atomic spacing in crystal copper is 3.6A, while the copper atomic radius is about 1.3A. Appendix E: How Thick is Surface Charge on a Metal? 346 The basic discussion above through (E.7) appears in Portis pp 162-164 (Chap 5, Sec 11). Portis then gives a small table of metal parameters and λD for copper is quoted as 0.59A, close to our result above. Thus, we come to the dramatic conclusion of this section: Fact : In our simple model, the thickness of the surface charge density below the surface of a conductor is incredibly small. For copper, it is l ess than the radius of one copper atom , and the general result applies to any metal. Thus, the surface charge decays away ri ght in the very first atomic layer of a metal. Fact : The thin layer of negative surface charge on the right plate in Fig E.1 above serves to neutralize the E field which would otherwise be present inside th e right conductor due to the positive charge on the surface of the left plate. One says th at the E field inside (and to the right of) the right plate is "screened" (killed off) by the negative surface charge layer on the right plate. This is of course the principle behind the ev er-popular Faraday Cage (note kids inside): Fig E.2 http://www.wonderwhizkids.com/resources/content/imagesv4/apupdate/phy sics/Electricity/conductors/Faraday_cage.jpg From Section 2.2, we found that the skin depth δ for copper at 100 GHz is about 0.2 microns which is 2x10-7m = 2000A. Even at this large frequency, the skin depth is still about 4000 times larger than the thickness of the surface charge layer. At 1 GHz this ratio is 40,000. Fact : Whereas surface current can exist "d eep" into the surface of a conductor, even when the skin effect is dominant, the surface charge can always be thought of as being exactly on the surface. Appendix F: Waveguides 347 Appendix F: Waveguides F.1 Discussion A transm ission line must have at least two distinct conductors to carry the TEM wave described in Section 3.7 and as illustrated in the figures there. Fo r a two conductor transmissi on line the surfaces of the conductors have a potential difference of amplitude V ≠ 0. A single wire cannot carry a TEM wave except in the sense of Section 2.1 where it acts as the center conductor of a coaxial cable with a far-distant re turn sheath. A TEM wave cannot propagate down the inside of a hollow pipe regardless of cross secti on shape since the continuous conductor cross section "shorts out" any possible V ≠ 0. In this document we have associated the TEM wave with the phrase "transmission line". but certainly a waveguide is a form of transmission line. Normal ly one associates the word "waveguide" with the TE and TM modes such waveguides carry. The usual form of a waveguide is in fact a hollow pipe, often of rectangular or circular cross section. However, it is possible for a 2 conductor transmission line to have TE and TM modes. In this Appendix we shall not pr esent a theory of waveguides since that is well done in Jackson and many other texts, but we would lik e to show that a transmission line made from two closely spaced parallel plates can carry waveguide mod es in addition to the TEM mode. We want to use this simple example to illustrate the notion that waveguide modes have lower cutoff frequencies whereas the TEM mode can operate all the way down to ω = 0 ( albeit in a very lossy manner). The terminology TEM (Transvers e Electric and Magnetic) means th at both the E and B fields are transverse, as shown in Figures 3.5 through 3.7. In r eality, we know there is a very small longitudinal E z field because E z is continuous at a conductor surface and we know J z = σEz just inside the conductor. This Ez field exists and has a cosine-like shape between the conductors, having the opposite direction at the second conductor. This field might be smaller than the transverse E field by a factor 10-4 as shown in (3.6.2). A TEM wave is very much like a plane wave with its transverse E and B fields, but the fields are distorted by the presence of the conductors. As Fig 3. 5 shows, this distortion is such that the Poynting vector E x B always points down the line (z direction), E and B are always perpendicular at any point [for sufficiently large ω, see (3.7.25)], and the E field lands perpendicularly on the conductors. The TE and TM modes have much more complicated field patterns. The waveguide modes are called TE (Transverse Electric) and TM (Transverse Magnetic). The nomenclature is a little confusing since both TE and TM waves generally have transverse E and B fields. The distinction is that the TE modes have no E z field, while the TM modes have no B z field. So TE means the E field is "transverse only". F.2 The TE waveguide modes for a parallel-plate transmission line We shall assu me (an "ansatz") that the entire E field is given by E(x,y,z) = E y(x) ej(ωt-kz) y^ ( F . 2 . 1 ) where we have our usual overloading of the symbol E. This field in the dielectric must satisfy the ω-domain wave equation (1.5.32) which says Appendix F: Waveguides 348 ( ∇2 + βd2) E = 0 . ( F . 2 . 2 ) Here βd is the usual Helmholtz parameter of the dielectric as in (1.5.1a), β d2 = μdεdω2 - jωμdσd = ω2μd ( εd - jσd/ω) = ω2μdξd ξd ≡ εd - jσd/ω . (1.5.1a) but in this Appendix we assume the dielectric is non-conducting so βd2 = ω2μdεd ( ≡ βd02 but we shall just call it βd2). Inserting our ansatz form (F.2.1) for E into (F.2.2) we find that (∂ x2 + ∂y2 + ∂z2+ βd2) Ey(x) ej(ωt-kz) = 0 or (∂ x2 + 0 +(-k2)+ βd2) Ey(x) = 0 . (F.2.3) Define γ 2 ≡ βd2-k2 ( F . 2 . 4 ) so that (∂ x2 + γ2) Ey( x ) = 0 ( F . 2 . 5 ) so Ey(x) = A sin( γx) + Bcos( γx) . (F.2.6) We now introduce our parallel plate transmissi on line (the gap is exaggerated in width) Fig F.1 Since we require E y = 0 at the two inner plate surfaces, we find that Appendix F: Waveguides 349 Ey(x) = A sin( γx) where sin( γd) = 0 => γd = mπ . (F.2.7) Thus, the parameter γ is quantized by the boundary conditions and we have E y(m)(x) = A sin( γmx) γm = m(π /d) m = 1,2,3.... (F.2.8) Suddenly we have "modes" la beled by m. The lowest non-vanishing mode has m = 1, and this is the mode shown in the figure. We find that the wave's wavenumber k is also quantized. From (F.2.4) we get k m = βd2 - γm2 . ( F . 2 . 9 ) In order for there to be a traveling wave ej(ωt-kz), we need k in (F.2.1) to be real, which requires that β d ≥ γm . ( F . 2 . 1 0 ) For a non-conducting dielectric one has βd = ωμdεd = ω /vd where v d is the light speeed in the dielectric. Recall from (1.1.29) that μ0ε0 = 1/c. So the above condition is ω/vd ≥ m(π/d) => ω ≥ m(π/d)vd so ω ≥ ωm ωm ≡ m(π/d)v = γm vd . ( F . 2 . 1 1 ) Thus the mth TE mode can only operate for ω above ωm, and as m increases the low end mode cutoff increases. For ω < ω1 there can be no TE action on this waveguide. The B fields for our TE mode can be obtained from the Maxwell curl E (1.6.19), B = (j/ω ) curl E = (j/ ω) [x^ (∂ yEz - ∂zEy) + y^ (∂zEx - ∂xEz) + z^ (∂xEy - ∂yEx)] = ( j / ω) [x^ (- ∂ zEy) + z^ (∂xEy) ] so then B x(m)(x) = (j/ω)(jk)Ey(x) = -(k m/ω) A sin(γ mx) (F.2.12) B z(m)(x) = (j/ω )∂xEy(x) = (j/ω) γm A cos(γ mx) . (F.2.13) Ey(m)(x) = A sin( γmx ) . (F.2.8) Appendix F: Waveguides 350 If A is real, then E y and Bx are real and in time phase, while B z is 900 out of phase. An attempt has been made to display all three field components in Fig F.1. To show that the waveguide mode outlined above is viable, we verify Maxwell's equations. Since the Maxwell curl E equation was used to obtain B, we need verify only the remaining three equations: div E = ∂ xEx + ∂yEy + ∂zEz = ∂yEy(x) = 0 (F.2.14) div B = ∂ xBx + ∂yBy + ∂zBz = ∂xBx + ∂zBz = -(k m/ω) γm Acos(γmx) - jkm (j/ω) γm A cos(γmx) = -(k m/ω) γm Acos(γmx) + km (1/ω) γm A cos(γmx) = 0 (F.2.15) Finally, curl B = x^ (∂yBz - ∂zBy) + y^ (∂zBx - ∂xBz) + z^ (∂xBy - ∂yBx) = + y^ (∂ zBx - ∂xBz) = y^ [ (-jkm )-(km/ω) A sin(γmx) + (j/ω ) γm2 A sin(γmx) ] = y^ [ (jkm2 /ω) A sin(γmx) + (j/ω) γm2 A sin(γmx) ] = y^ [ km2 + γm2 ] j (A/ω)sin(γmx) = y^ βd2 j (A/ω)sin(γmx) . (F.2.16) According to (1.6.18) curl H(x,ω) = jωD(x,ω) + J (x,ω) (1.6.18) with J = 0 , D = εE and H = B/μ we should have curl B = jωεμ E ( F . 2 . 1 7 ) or β d2 j (A/ω)sin(γmx) = jωεμ A sin(γmx) // (F.2.16) and (F.2.8) or βd2 (1/ω) = ωεdμd or βd2 = ω2εdμd which is (1.5.1b) quoted above. T hus we have shown that our TE waveguide modes satisfy all four of Maxwell's equations. Although the TE and TEM modes are both "transverse electric", there is a significant difference in the E field pattern. In TEM the E field lines run from one conductor to the other so that the line integral of E generates the potential difference V, as shown in Fig 3.5. The E field lines are "sourced by" (or "create") the surface charge on the conductors. In the TE mode of Fig F.1, the E field is still transverse but is Appendix F: Waveguides 351 parallel to the conductors so the line integral of E between the conductors gives V = 0. These E lines are not sourced by charges on the conductors but are more like the E field lines in a free-space light wave. Comments: 1. The parallel plate transmission line also has TM waveguide modes, and the cutoff frequencies are the same as for the TE mode. 2. A rectangular waveguide mode has two quantized integers and the cutoff frequency is then a function of both these integers. For TM the E z field will have sine behavior in both x and y directions. 3. The obvious boundary condition is that E t = 0 at the walls, while a less obvious condition is that Bn = 0 at the walls [ see (3.7.17) ]. Notice in our example that B x(m)(x) = 0 at the walls and this is a normal B field. 4. Waveguide problems are normally dealt with usi ng the Helmholtz equation for the E and B fields, whereas the TEM transmission line problem is more easily dealt with using potentials φ and Az. 5. We have dealt above with an ideal waveguide. In real waveguides the fields E and B penetrate distance δ (skin depth) into the walls and ge nerate ohmic losses causing the wave to be damped. The same thing of course also happens for the transmission line TEM mode. Reader Exercise: Make a 3D vector plot of the E and B fields for Fig F.1, and also plot the Poynting vector S = E x B and compare with the TEM wave pattern. Except at the center, in addition to S z there seems to be an S x component suggesting a transverse power flow distribution in addition to the expected longitudinal power flow. Appendix G: The DC vector potential of a round wire 352 Appendix G: The DC vector potential of a round wire carrying a uniform current In this proble m, an isolated, infin itely-long and z-aligned round wire ( μ2,ε2,σ2, radius a ) carries a current I. The wire is immersed in an infinite dielectric medium ( μ1,ε1,σ1). We begin for general ω, but quickly go to the DC limit ω = 0. We wish to calculate the vector potential A of this wire both inside and outside . Section G.1 sets up the problem, makes some ansatz assumptions, and then ends up with a 2D Poisson equation for the potential which is ∇2 2D Az(r) = - [Iμ2/(πa2)] θ(r≤a). Section G.2 directly solves this Poisson equation for the potential A z(r). The solution is required to meet two boundary conditions at r = a. Section G.3 very quickly computes this same A z(r) using Ampere's Law with the same boundary conditions and obtains the same result found in Section G.2. Section G.4 laboriously obtains the same A z(r) result using the 2D Helmholtz integral (which in this case is really just a Poisson integral). This serves as a prototype case for dealing with such integrals, so much detail is provided. It is found that for μ1 ≠ μ2 homogenous terms must be added to the Helmholtz integral in order to meet the boundary conditions. Section G.5 comments on the solution for A z at low frequencies. G.1 Setup and Assumptions The ω-domain Helm holtz wave equation for A using the King gauge is given by (1.5.4), ( ∇2 + β12)A = - μ2J2 β12 = ω2μ1 ξ1 ξ1 ≡ ε1 - jσ1/ω (G.1.1) div A = jωμ1ξ1φ . // King gauge [ 1 = dielectric, 2 = wire ] (G.1.2) We take a uniform prescribed current inside the wire ( assume low frequency), J2 = [I/(πa2)] z^ ( G . 1 . 3 ) so the Helmholtz wave equation reads ( ∇2 + β12)A(x) = - [Iμ2/(πa2)] θ(x2+y2 < a) z^ where ∇2 is the vector Laplacian and where θ(B) = 1 if B is true, else 0. Using Cartesian components, this says ( ∇ 2 + β12)Ax(x) = 0 ( ∇2 + β12)Ay(x) = 0 ( ∇2 + β12)Az(x) = - [Iμ2/(πa2)] θ(x2+y2 < a) where in these three equations ∇2 is the scalar Laplacian. We shall seek a solution in which both A x and Ay vanish. In this case we have Appendix G: The DC vector potential of a round wire 353 A(x) = Az(x) z^ . We assume a very low frequency ω for which we know any longitudinal wave that might be going down the wire has a very long wavelength. We then ignore z variations in A z to write A(x) = Az(x,y) z^ . ( G . 1 . 4 ) In this case, one finds that ∇2Az = ∇2 2D Az and then our only equation of interest is this: ( ∇2 2D + β12) Az(x,y) = - [I μ2/(πa2)] θ(x2+y2 < a) . // ∇2 2D = ∇2 - ∂z2 (G.1.5) We now take ω→0 to get ∇2 2D Az(x,y) = - [I μ2/(πa2)] θ(x2+y2 < a2) which is just a 2D Poisson equation with a constant source limited to a region of space. At this point we are free to replace x,y with polar coordinates r, θ, so we have for r in the range (0, ∞), ∇2 2D Az(r,θ) = - [Iμ2/(πa2)] θ(r<a) . // θ(r<a) = Heaviside θ(a-r). (G.1.6) From B = curl A in cylindrical coordinates we find that, since only A z is non-vanishing, B = curl A = r^ [ r-1∂θAz - ∂zAθ] + θ^ [∂zAr - ∂rAz] + z^ [ r-1∂r(rAθ) - r-1∂θAr ] = r^ [ r -1∂θAz] + θ^ [- ∂rAz] . Since we expect the magnetic fi eld lines to be entirely in the θ^ direction, we are led to make the assumption that ∂θAz = 0 and then the problem is this ∇2 2D Az(r) = - [Iμ2/(πa2)] θ(r<a) B = Bθθ^ with B θ = - ∂rAz . (G.1.7) If we can find a solution, then our ansatz assumptions that A x = Ay = 0 and B = Bθθ^ are justified. G.2 Direct solution for A z(r) from the differential equation Using ∇2 2D in polar coordinates our ODE (G.1.7) reads (1/r)∂r(r∂rAz(r)) = - [Iμ2/(πa2)] θ(r≤a) or ∂ r(r∂rAz(r)) = - [Iμ2/(πa2)] r θ(r≤a) or r A z"(r) + A z'(r) = - [Iμ2/(πa2)] r θ(r≤a) . (G.2.1) Appendix G: The DC vector potential of a round wire 354 The two regional differential equations are then r A z"(r) + A z'(r) = 0 r>a region 1 r Az"(r) + A z'(r) = - [Iμ 2/(πa2)] r r<a region 2 . (G.2.2) The general-form solutions to these ODE's are, A z(r) = C ln(r) + D r>a region 1 Az(r) = - [I μ2/(4πa2)] r2 + E ln(r) + F r<a region 2 (G.2.3) where there are 4 constants to be determined. For r>a the functions 1 and lnr are the well-known atomic forms (harmonic elements) for the 2D Laplace equation for situations of azimuthal symmetry. The first term in region 2 (G.2.3) is the particular solution of region 2 (G.2.2) to which we have added a possible homogeneous solution E ln(r) + F. In order that A z(r) be finite at r = 0, we must set E = 0. For very large r the round wire looks like a line source and we know the solution of that problem. Using Ampere's Law that 2 πrHθ = I we find ( recall that B θ = - ∂rAz) H θ = [I/2π ](1/r) => B θ = μ1 [I/2π](1/r) => A z = - [I μ1/2π] ln(r) + D' . r >> a Comparing this solution to our r>a round wire solution we conclude that D' = D and C = -[ μ 1I/2π] . There are still two unknown constants D and F : Az(r) = -[Iμ1/2π] ln(r) + D r > a region 1 Az(r) = - [Iμ2/(4πa2)] r2 + F r < a region 2 . (G.2.4) The potential A z(r) always has an additive constant which we are free to specify and which affects nothing. We shall choose the zero point of A z(r) by setting D = 0 arbitrarily. This means that A z(r) has the simple form K ln(r) for r>a, and that choice implies that A z(a) = - [Iμ 1/2π] ln(a), so we have in effect specified A z(r) on the wire surface to be this value. Notice for future reference that ∂ rAz(r) = - [Iμ1/2π] (1/r) r > a region 1 ∂rAz(r) = - [Iμ2/(2πa2)] r r < a region 2 . (G.2.5) Now, since our prescribed current J 2 does not specify a free surface current K z on the round wire surface, which would have the form Js,surface = Kzfree δ(a-r), we conclude that there is no free surface current on the round wire surface; there is only the bulk volume current J z = I/(πa2)θ(r<a). Therefore the boundary condition (1.1.46) applies (though now in polar coordinates) and we conclude that Appendix G: The DC vector potential of a round wire 355 (1/μ1) [(∂rAz)(a)]1 = (1/μ2) [(∂rAz)(a)]2 . In addition, we shall require that A z itself be continuous at the boundary, so here are our two boundary conditions of interest (superscript 1 means region 1 which is r>a), [Az(a)]1 = [Az(a)]2 . (1/μ1) [(∂rAz)(a)]1 = (1/μ2) [(∂rAz)(a)]2 . ( G . 2 . 6 ) We now require that both these boundary conditions be met by the A z expressions of (G.2.4), [A z(a)]1 = [Az(a)]2 // (G.2.6) repeated (1/μ1) [(∂rAz)(a)]1 = (1/μ2) [(∂rAz)(a)]2 or -[Iμ1/2π] ln(a) = - [I μ2/(4πa2)] a2 + F // insert expressions, set r = a (1/μ1){- [Iμ1/2π] (1/a)} = (1/ μ2){ - [Iμ2/(2πa2)]a } or -[Iμ1/2π] ln(a) = - [I μ2/(4π)] + F // simplify 1 = 1 The second boundary condition is thus met automatically by our solution. The first says F = [Iμ 2/(4π)] - [Iμ1/2π] ln(a) and so the solution is then A z(r) = - [Iμ1/2π] ln(r) r > a region 1 Az(r) = - [Iμ2/(4πa2)] r2 + [Iμ2/(4π)] - [Iμ1/2π] ln(a) r < a region 2 or A z(r) = - [Iμ1/2π] ln(r) r > a region 1 Az(r) = - [Iμ2/(4πa2)] (r2-a2) - [Iμ1/2π] ln(a) r < a region 2 . (G.2.7) This then is the complete solution to the problem for the round wire, ∇2 2D Az(r) = - [Iμ2/(πa2)] θ(r≤a) B = Bθθ^ with B θ = - ∂rAz . (G.1.7) where the potential is "pinned" by the requirement that A z = K ln(r) for r > a. We may now compute the B field from our potential solution (G.2.7), Bθ = - ∂rAz = [Iμ1/2π]∂rln(r) = [I μ1/2π](1/r) r > a region 1 Bθ = - ∂rAz = [Iμ2/(2πa2)] r = [Iμ2/2π](r/a2) r < a region 2 . (G.2.8) Appendix G: The DC vector potential of a round wire 356 Comment : Although only μ 2 appears in the differential equation (G.1.7), once the equation is properly solved with attention to boundary conditions, we find that μ1 appears in B θ in region 1, while μ 2 appears in Bθ in region 2! This is the ma in point of this Section G.2. G.3 Instant solution for A using Ampere's Law and computation of J m For r > a Ampere's law (1.1.37) (converted to ω space with ω = 0) says 2 πrHθ = I so Hθ = I/(2πr) => B θ = μ1I/(2πr) [ 1 = dielectric ] r > a r e g i o n 1 ( G . 3 . 1 ) => - ∂ rAz = μ1I/(2πr) => A z = - [μ1I/(2π)] ln(r) + D . For r < a Ampere's law says 2 πrH θ = I(πr2/πa2) [ the "current enclosed" ] Hθ = I(r2/a2)1/(2πr) = I r/(2 πa2) => B θ = [Iμ2/2π](r/a2) [ 2 = wire ] r < a r e g i o n 2 ( G . 3 . 2 ) => - ∂rAz = [Iμ2/2π](r/a2) => A z = - [Iμ2/4π](r2/a2) + E . We then set D = 0 to get A z = K ln(r) for r > 0, as done previously, and then we must match at r = a : - [μ 1I/(2π)] ln(a) = - [I μ2/4π] + E => E = [Iμ 2/4π] - [μ1I/(2π)] ln(a) so our potential solution is then A z(r) = - [μ1I/(2π)] ln(r) r>a Az(r) = - [Iμ2/4π](r2/a2) + { [Iμ2/4π] - [μ1I/(2π)] ln(a) } r<a or A z(r) = - [μ1I/(2π)] ln(r) r>a Az(r) = - [Iμ2/4πa2]r2 + [Iμ2/4π] - [μ1I/(2π)] ln(a) r<a or Az(r) = - [μ1I/(2π)] ln(r) r>a Az(r) = - [Iμ2/4πa2](r2-a2) - [μ1I/(2π)] ln(a) r<a (G.3.3) This result agrees with result (G.2.7) of the previous section. The Magnetization Current It was mentioned in Section G.2 that there is no free surface current K zfree at the wire surface. There is in fact a bound magnetization current on this surface and inside the wire as well. Luckily, our Helmholtz equation only "sees" conduction currents so we don't ha ve to worry about the magnetization currents. The magnetization current density is given by Jm = curl M where M = [μ /μ0- 1] H as shown in (1.1.20-23). Appendix G: The DC vector potential of a round wire 357 Here then is the calculation of Jm : Mθ1 = [ μ1/μ0 - 1] Hθ1 = [ μ1/μ0 - 1] (I/2πr) r > a Mθ2 = [ μ2/μ0 - 1] Hθ2 = [ μ2/μ0 - 1] (Ir/2πa2) r < a or Mθ(r) = [ μ1/μ0 - 1] (I/2πr) θ(r>a) + [ μ2/μ0 - 1] (Ir/2π a2)θ(r<a) // θ(r>a) = θ(r-a) ∂rMθ = - [ μ1/μ0 - 1] (I/2πr2) θ(r>a) + [ μ1/μ0 - 1] (I/2πr) δ(r-a) + [ μ2/μ0 - 1] (I/2πa2) θ(r<a) + [ μ2/μ0 - 1] (Ir/2π a2) )[ -δ(r-a)] = - [ μ 1/μ0 - 1] (I/2πr2) θ(r>a) + [ μ2/μ0 - 1] (I/2πa2)θ(r<a) + [ μ1/μ0 - μ2/μ0] (I/2πa) δ(r-a) J mz = r-1∂r{rMθ} = r-1[ Mθ + r∂rMθ] = r-1Mθ + ∂rMθ = [ μ 1/μ0 - 1] (I/2πr2) θ(r>a) + [ μ 2/μ0 - 1] (I/2πa2)θ(r<a) - [ μ1/μ0 - 1] (I/2πr2) θ(r>a) + [ μ2/μ0 - 1] (I/2πa2)θ(r<a) + [ μ1/μ0 - μ2/μ0] (I/2πa) δ(r-a) = [ μ 2/μ0 - 1] (I/πa2)θ(r<a) + [ μ1/μ0 - μ2/μ0] (I/2πar) δ(r-a) . (G.3.4) constant inside wire surface current It is the discontinuity of M θ(r) at the wire surface r = a which creates the surface current term in J mz. The simplest possible case to consider is th e boundary at x = 0 between two half spaces of μ1 and μ2 assuming there exists a uniform constant H y field everywhere. In this case, one would have M y1(x) = [ μ 1/μ0 - 1] Hy θ(x) // H y is continuous at the boundary by (1.1.42) My2(x) = [ μ 2/μ0 - 1] Hy θ(-x) (G.3.5) ∂ xMy(x) = [ μ1/μ0 - μ2/μ0] Hy δ(x) J mz = [curl M] z = ∂xMy = [ μ1/μ0 - μ2/μ0] Hy δ(x) . (G.3.6) so here there is only a surface ma gnetization current and no bulk magnetization current on either side. G.4 Solution for A z using the 2D Helmholtz Integral This method of finding A z is technically more difficult than the first two methods shown in Section G.2 (solving the ODE and adding a homogeneous solution to meet the boundary conditions) and Section G.3 (instant Ampere's Law solution). The method is impor tant because our entire Chapter 4 is based on using Helmholtz integrals to develop the theory of transmissi on lines, and this is one Helmholtz integral that we can actually compute without too much effort. An important result we find is that, when μ1≠μ2, the Helmholtz integral by itself does not supply the comple te solution, and one must add in some amount of homogeneous solution of ∇2 2D Az(r) = 0 to meet the required boundary conditions at r=a. We really have a Poisson integral since β12 = 0, but the full Helmholtz integral works the same way so we keep referring to it as a Helmholtz integral. Appendix G: The DC vector potential of a round wire 358 Recall from above: ∇ 2 2D Az(r) = - [Iμ2/(πa2)] θ(r≤a) B = Bθθ^ with B θ = - ∂rAz . (G.1.7) Using the 2D free-space Green's function (propagato r) as reviewed in Appendix I equation (I.1.6), g(x|x') = (1/2 π) ln(1/R) = - (1/4 π) ln(R2) R = R = | x-x'| , (G.4.1) we may write the particular solution to (G.1.7) as the following "Helmholtz" integral [see (I.1.8)] -AzH(r) = ∫dS' [(1/4π) ln(R2) ] [Iμ2/(πa2)] θ(r≤a) = ∫0 a r' dr' ∫-π π dθ' [(1/4π) ln(r2+ r'2- 2rr'cos(θ-θ'))] [Iμ2/(πa2)] = ( 1 / 4 π) [Iμ2/(πa2)] ∫0 a r' dr' [ ∫-π π dθ' ln(r2+ r'2- 2rr'cos(θ-θ')) ] = μ2I 4π2a2 ∫0 a r' dr' [ 2 Q(r',r) ] (G.4.2) where Q(r',r) ≡ (1/2) ∫-π π dθ' ln(r2+ r'2- 2rr'cos(θ-θ')) = (1/2) ∫-π π dθ" ln(r2+ r'2- 2rr'cos(θ")) = ∫0 π dx ln(r2+ r'2- 2rr'cosx) . (G.4.3) But we have already computed this A zH(r) in Appendix B where it was called A z(c)(r,θ), see (B.7.3) and (B.7.4). We may therefore borrow the so lution (B.7.7) to obtain the results, AzH(r>a) = - μ2I πa2 { (1/2) a2lnr } = - μ2I 2π lnr AzH(r<a) = μ2I 2πa2 { (a2-r2)/2 - a2lna } . (G.4.4) The derivatives are ∂ rAzH(r>a) = - μ2I 2πr ∂rAzH(r<a) = - μ2I 2πa2 r . ( G . 4 . 5 ) Appendix G: The DC vector potential of a round wire 359 Recall the two boundary conditions, [Az(a)]1 = [Az(a)]2 . (1/μ1) [(∂rAz)(a)]1 = (1/μ2) [(∂rAz)(a)]2 . (G.2.6) For our particular Helmholtz integral A zH(r) we evaluate these boundary conditions to find - μ2I 2π lna = μ2I 2πa2 { (a2-a)/2 - a2lna } (1/μ1) [- μ2I 2πa ] = (1/μ2)[- μ2I 2πa2 a] or 1 = 1 (μ2/μ1) = 1 . (G.4.6) Thus, only in the case μ1 = μ2 does the Helmholtz particular solution meet both boundary conditions. If μ1 ≠ μ2, we must add to the particular solution some amount of homogeneous solution of ∇2 2D Az(r,θ) = 0. So we then write generally, Az(r) = AzH(r) + Azhomo(r) (G.4.7) where we know that A zhomo(r) can only have terms α + β ln r. We then write for the two regions Az(r) = -μ2I 2π ln(r) + α + β l n r r > a Az(r) = μ2I 2πa2 { (a2-r2)/2 - a2lna } + α ' + β ' lnr r<a . (G.4.8) As earlier, we choose the zero point for A z(r) by requiring that the large r behavior be K ln(r) without a constant added, which then means α = 0. And for r<a we must have β' = 0 to be finite at r = 0. So Az(r) = -μ2I 2π ln(r) + β l n r r > a Az(r) = μ2I 2πa2 { (a2-r2)/2 - a2lna } + α ' r<a (G.4.9) -∂ rAz(r) = [ μ2I 2π - β ] ( 1 / r ) r > a -∂rAz(r) = μ2I πa2 (2r) r<a . (G.4.10) The boundary conditions are then Appendix G: The DC vector potential of a round wire 360 [Az(a)]1 = [Az(a)]2 (1/μ1) [(-∂rAz)(a)]1 = (1/μ2) [(-∂rAz)(a)]2 (G.2.6) or -μ 2I 2π ln(a) + β lna = μ2I 2πa2 { (a2-a2)/2 - a2lna } + α ' (1/μ1) [μ2I 2π - β](1/a) = (1/ μ2) μ2I πa2 (2a) or [- Iμ2/(2π) + β] lna = - I μ2/(2π) lna + α ' (1/μ1)[ Iμ2/(2π) - β](1/a) = I/(2 πa) // simplify or β lna = α ' Iμ2/(2π) - β = μ1I/(2π) // simplify some more so we find that β = I/(2π) (μ 2-μ1) α' = I/(2π) (μ2-μ1) l n a . ( G . 4 . 1 1 ) The full solution is then Az(r) = [- Iμ 2/(2π) + β ] l n r r > a Az(r) = - Iμ 2/(πa2) { (1/4) (r2-a2) + (1/2) a2lna } + α ' r < a or A z(r) = [- Iμ 2/(2π) + {I/(2π) ( μ2-μ1)}] lnr r>a Az(r) = - Iμ 2/(πa2) { (1/4) (r2-a2) + (1/2) a2lna } + { I/(2 π) ( μ2-μ1) lna } r<a or A z(r) = I/(2 π) [- μ2 + ( μ2-μ1) ] l n r r > a Az(r) = - Iμ 2/(πa2) (1/4) (r2-a2) - Iμ2/(πa2) (1/2) a2lna + I/(2 π) ( μ2-μ1) lna r<a or A z(r) = - (I/(2 π) [μ1] l n r r > a Az(r) = - Iμ 2/(πa2) (1/4) (r2-a2) - Iμ1/(π) (1/2) lna r<a or A z(r) = - [Iμ1/2π] l n r r > a Az(r) = - [Iμ2/(4πa2)] (r2-a2) - [Iμ1/2π] lna r<a . (G.4.12) This result matches the results (G.2.7) and (G.3.3) of the previous two methods and then gives the B field solution, Bθ = - ∂rAz = [Iμ1/2π]∂rln(r) = [I μ1/2π](1/r) r > a region 1 Bθ = - ∂rAz = [Iμ2/(2πa2)] r = [Iμ2/2π](r/a2) r < a region 2 . (G.2.8) Appendix G: The DC vector potential of a round wire 361 G.5 Comments on the low frequency solution for A z In Chapter 2 we compute the E and B fields inside a round wire operating at frequency ω. The results are rather complicated and involve special Bessel functi ons called Kelvin functions. The vector potential was not used in that Chapter. Here we consider computing A z using the true Helmholtz integral rather than its Poisson approximation, and see how the derived resu lts might compare with the Chapter 2 results. Comments: 1. For sufficiently low frequencies (the transmissi on line limit) we imagine that the ansatz assumptions we made in Section G.1 are still pretty good. The curre nt distribution will be nearly uniform. There will likely be some small A x and Ay fields which we can ignore, and we still assume roughly that B = Bθθ^ with B θ = - ∂rAz and that we can ignore the z-dependence of A z, though we know it must vary some small amount in order to have a long- λ wave passing down the wire. Therefore, our problem is basically (G.1.5) for small β12, ( ∇2 2D + β12) Az(r) = - [Iμ2/(πa2)] θ(r<a) B = Bθθ^ with B θ = - ∂rAz . (G.5.1) 2. Since ∇2 2D = (1/r)∂r(r∂r), one could write out the above differe ntial equation and repeat the work of section G.2 above. The resulting B field obtained from B θ = - ∂rAz should then agree with the low frequency limit of (2.2.25) which applies inside the round wire, Bθ(r) = B θ(a) J1(β1r) J1(β1a) . (2.2.25) That low-frequency limit is Bθ(r) ≈ Bθ(a) r - β12r3/8 a - β12a3/8 β12 = ω2μ ξ1 . (G.5.2) Certainly as ω → 0 (so β1→ 0) the result B θ(r) = B θ(a)(r/a) agrees with (G.2.8). 3. The 2D free-space Helmholtz propagator is shown in (I.1.7) to be g(x|x') = (j/4) H 0(1)(β1R) R = R = | x-x' | where H 0(1) is a Hankel function. Thus, we may write the particular solution to (G.5.1) as the following Helmholtz integral [ see (I.1.9) ] , AzH(r) = ∫dS' [(j/4) H 0(1)(kR) ] [Iμ2/(πa2)] θ(r≤a) R = R = | x-x'| = [I μ2/(πa2)] (j/4) ∫0 a r' dr' ∫-π π dθ' H0(1)(β1r2+ r'2- 2rr'cos(θ-θ') ) . (G.5.3) Appendix G: The DC vector potential of a round wire 362 The dθ' integral is actually doable with this resu lt (making use of GR7 p 726 6.684 1 and 2) ∫-π π dθ' H0(1)(β1r2+ r'2- 2rr'cos(θ-θ') ) = (1/2) { π J0(β1r) H0(1)(β1r')θ(r'>r) + π J0(β1r') H0(1)(β1r)θ(r'<r) } . (G.5.4) The two dr' integrals can then be done (using GR7 p 629-630 Section 5.5) with the final result A zH(r) = [Iμ 2/(πa2)](j/4)2π (1/β1) * { J 0(β1r) θ(a>r) [a H 1(1)(β1a) - r H1(1)(β1r) ] + H 0(1)(β1r) ⎩⎨⎧ r J1(β1r) r<a a J1(β1a) r>a } . (G.5.5) We leave it to the reader to determine the small β1 limit of this result and see if the resulting B θ = - ∂rAz agrees with (G.5.2) after homogeneous solutions are added to match boundary conditions. Remember that we only expect this result to be meaningful for low ω since we have assumed the uniform current distribution of (G.1.3). If one makes the small-argument approximation H 0(1)(x) ≈ (2j/π )ln(x) directly in (G.5.3), the integral replicates the Poisson result (G.4.2), so more expa nsion terms would be needed for this approach. Appendix H : Poisson and He lmholtz Propagators in 3D 363 Appendix H: Poisson and Helmholtz Propagators in 3D Note : Appendix I deals with these propagators in 2D rather than 3D. __________________________________________________________________________________ H.1 Overview and Meaning of Free-Space Propagators This appendix proves the following Facts: Fact 1 : - ∇ 2[1/4πr] = δ(r) (H.2.1) (H.1.1) Fact 2 : - ∇ 2[h(r)/r] = 4 π h(0) δ (r) - h"(r)/ r (H.3.1) (H.1.2) Fact 3 : - (∇ 2+k2) (e-jkr/4πr) = δ(r) (H.3.5) (H.1.3) Throughout, ∇2 is the usual 3D Laplacian operator ∇2 = ∂x2 + ∂y2 + ∂z2. In the first and last results above, if one replaces r → r-r' (a simple translational shift of origin) ones finds -∇ 2[1/4πR] = δ(r-r') R = | r - r' | (H.1.4) - (∇2+k2) [e-jkR/4πR] = δ (r-r') δ(r-r') = δ(x-x') δ (y-y') δ(z-z') (H.1.5) The quantities in brackets are known as free-space Green 's Functions (Green F unctions) or propagators, or as "fundamental solutions": 1/4πR = the Poisson 3D free-space propagator (H.1.6) e -jkR/4πR = the Helmholtz 3D free-space propagator . (H.1.7) The last item above is the ω-domain 3D Helmholtz propagator, where k 2 = ω2με. See (A.7.4) for a discussion of the time domain version of this propaga tor which is the 3D wave equation propagator. The significance of these propagators is the following: -∇ 2 f(x) = s( x) => f( x) = ∫d3x' [1/4πR] s( x') + homogeneous solutions T h e P o i s s o n E q u a t i o n ( H . 1 . 8 ) - (∇2+k2) f(x) = s( x) => f( x) = ∫d3x' [e-jkR/4πR] s( x') + homogeneous solutions T h e H e l m h o l t z E q u a t i o n ( H . 1 . 9 ) The equations on the left are inhomogeneous partial differential equations driven by source function s( x). If one is careful to include in s( x) all source contributions (such as those on boundary surfaces), one generally does not have to add any homogeneous solutions on the right. A homogeneous solution refers to Appendix H : Poisson and He lmholtz Propagators in 3D 364 -∇2 fh(x) = 0, for example. The solutions shown on the ri ght above can be instantly verified as follows: f(x ) = ∫d3x' [1/4πR] s( x') + fh(x) -∇2 f(x) = ∫d3x' (-∇2 [1/4πR] ) s( x') -∇2 fh(x) = ∫d3x' δ(r-r') s(x') - 0 = s( x) (H.1.10) and similarly for - ( ∇2+k2) f = g. A "free space" Green's Function g F in general is a solution of Lr gF(r, r') = δ (r-r'), g F(r, r') → 0 as r → ∞ ( H . 1 . 1 1 ) where L r is some differential operator. The condition on the right says g F must vanish on the Great Sphere. More generally one can define a full Green's function by, Lr g(r, r') = δ(r-r'), g( r, r') = 0 for r on some closed surface enclosing a region of interest (H.1.12) This non-free-space Green's function is briefly discussed in the text surrounding (1.5.11). George Green (1793-1841), by the way, was an English grain miller (his day job). Looking at f( x) = ∫d3x' [1/4πR] s( x') = ∫ gF(x,x') [s( x') d3x'], one can say that the kernel Green's Function g F(x,x') "propagates" a tiny piece of "source" [s( x')d3x'] from location x' to location x so that the solution f( x) is then a sum of all such propagated c ontributions as the source ranges over the entire volume of interest, which for us is all 3D space where the source is non-vanishing. See Fig 1.6. __________________________________________________________________________________ H.2 Derivation of Fact 1: - ∇2[1/r] = 4πδ(r) ( H . 2 . 1 ) Proof: Let volume V be all of 3D space. Carve out from V a small spherical cavity of radius a centered at r = 0. If we call this spherical volume V a and then V' = V - V a is the original volume with the spherical cavity carved out: Fig H.1 In order to show that some function g( r) = δ(r), one has to show that Appendix H : Poisson and He lmholtz Propagators in 3D 365 lima→0 ∫V' dV g( r) = 0 ( H . 2 . 2 a ) lima→0 ∫Va dV g( r) = 1 . ( H . 2 . 2 b ) This is basically the definition of δ(r). Since δ(r) has units L-3, g(r) = g( r) has units L-3. Our candidate function of interest is g(r) = - (1/4 π) ∇ 2[ 1 / r ] . ( H . 2 . 3 ) Using ∇ 2 in spherical coordinates acting on a function of r, one finds that, since ∂r(1) = 0, ∇ 2[1/r] = (1/r2)∂r(r2∂r) [1/r] = 0 r > 0 (H.2.4) so that g(r) = - (1/4 π) ∇2[1/r] = 0 r > 0 . (H.2.5) Thus, condition (H.2.2a) is trivially satisfied since r > 0 everywhere in volume V'. It remains to verify condition (H.2.2b). Consider th e integral appearing on the left side of (H.2.2b) ∫Va dV g(r) = - (1/4 π) ∫Va dV ∇2[1/r] = - (1/4 π) ∫Va dV ∇ • ∇[1/r] . (H.2.6) The divergence theorem (1.1.30) says, ∫V dV div F = ∫S dS • F ( H.2.7) where V is any closed volume whose surface is S, and d S points out. Using V = V a and F = ∇[1/r] = r^∂r(1/r) = -r-2r^ we find that LHS (H.2.7) = ∫Va dV div ∇ [1/r] = ∫Va dV ∇2[1/r] = ∫Va dV [-4πg(r)] = -4π ∫Va dV g(r) RHS (H.2.7) = ∫S dS • ∇[1/r] = ∫dΩ [a2 r^] • ∇[1/r]|r=a = ∫dΩ[a2 r^]• [-a-2r^] = -4π which tells us that ∫Va dV g(r) = 1 for any a. Thus, Appendix H : Poisson and He lmholtz Propagators in 3D 366 lima→0 ∫Va dV g(r) = 1 and we have then verified (H.2.2b). Th erefore we conclude that the candidate g( r) of (H.2.3) is in fact the same as δ(r) so - (1/4 π)∇2[1/r] = δ (r) ( H . 2 . 8 ) or ∇ 2[1/r] = - 4 πδ(r) ( H . 2 . 9 ) w h i c h i s ( H . 2 . 1 ) . QED __________________________________________________________________________________ H.3 Derivation of Fact 2: ∇2[h(r)/r] = - 4 π h(0) δ (r) + h"(r)/ r (H.3.1) Proof : Start with this vector identity, ∇2(φψ) = φ ∇2ψ + ψ∇2φ + 2 ∇φ • ∇ψ . ( H . 3 . 2 ) This identity is valid in any number of dimensions (implied sum on i from 1 to N) , ∂ i2(φψ)= ∂i[ (∂iφ)ψ + ψ(∂iφ)] = (∂i2φ)ψ + (∂iφ) (∂iψ) + φ (∂i2ψ) + (∂iφ) (∂iψ) . So apply (H.3.2) to the case φ = h and ψ = r -1, ∇2(h r-1) = h ∇2(r-1) + r-1∇2h + 2 ∇ h • ∇(r-1) = - h 4 π δ(r) + r-1∇2h + 2 [ h' r^ • (-r-2) r^ ] // using (H.2.9) = - 4 π h(0) δ (r) + r-1∇2h - 2 r-2 h'(r) . (H.3.3) Algebra shows that, using spherical coordinates, ∇2h = (1/r2)∂r(r2∂r)h(r) = h"(r) + (2/r)h'(r) (H.3.4) so then ∇ 2(h r-1) = - 4π h(0) δ (r) + r-1 [h"(r) + (2/r)h'(r) ] - 2 r-2 h'(r) = - 4 π h(0) δ (r) + h"(r)/ r w h i c h i s t h e c l a i m o f ( H . 3 . 1 ) . QED Appendix H : Poisson and He lmholtz Propagators in 3D 367 Fact 3: - ( ∇2+k2) (e-jkr/4πr) = δ(r) ( H . 3 . 5 ) This Fact is just an application of Fact 2 to the case h(r) = e-jkr : h = e -jkr h(0) = 1 h' = -jk e-jkr h" = -k2 e-jkr ∇2[h(r)/r] = - 4 π h(0) δ (r) + h"(r)/ r (H.3.1) so ∇2(e-jkr/r) = - 4π 1 δ(r) + [-k2 e-jkr ] / r = - 4 πδ(r) - k2(e-jkr/r) Thus, ( ∇2+k2) (e-jkr/r) = - 4πδ(r) or - (∇ 2+k2) (e-jkr/4πr) = δ(r) as claimed. Appendix I : Poisson and He lmholtz Propagators in 2D 368 Appendix I: Poisson and Helmholtz Propagators in 2D Note : Appendix H deals with these propagators in 3D ra ther than 2D. Sections I.1 and I.2 below are basically "cut, paste and edit" versions of Sections H.1 and H.2, and we ha ve made equation numbers match. However, Section I.3 is something new since it involves a "special function". __________________________________________________________________________________ I.1 Overview and Meaning of Free-Space Propagators This appendix proves two Facts: (H 0(1) is a Hankel function ) Fact 1 : - ∇2[ln(1/r)/2π] = δ (r) (I.2.1) (I.1.1) Fact 2 : - (∇ 2+k2) [(j/4) H 0(1)(kr)] = δ (r) . (I.3.1) (I.1.2) Throughout this Appendix, ∇2 is the usual 2D Laplacian operator, ∇ 2 = ∂x2 + ∂y2. and δ(r) = δ(x) δ(y) . (I.1.3) In the two Facts above, if one replaces r → r-r' (a simple translational shift of origin) ones finds -∇2[ln(1/R)/2 π] = δ(r-r') R = | r - r' | (I.1.4) - (∇ 2+k2) [(j/4) H 0(1)(kR)] = δ(r-r') δ(r-r') = δ(x-x') δ (y-y') . (I.1.5) The quantities in brackets are known as free-space Green 's Functions (Green F unctions) or propagators, or as "fundamental solutions" : 1 2π ln(1/R) = the Poisson 2D free-space propagator (I.1.6) (j/4) H 0(1)(kR) = the Helmholtz 2D free-space propagator . (I.1.7) The last item above is the ω-domain 2D Helmholtz propagator, where k 2 = ω2με. See (A.7.7) for a discussion of the time domain version of this propaga tor which is the 2D wave equation propagator. The significance of these propagators is the following: -∇ 2 f(x) = s( x) => f( x) = ∫d2x' [ln(1/R)/2π ] s(x') + homogeneous solutions The Poisson Equation (I.1.8) - (∇2+k2) f(x) = s( x) => f( x) = ∫d2x' [(j/4) H 0(1)(kR)] s( x') + homogeneous solutions The Helmholtz Equation (I.1.9) Appendix I : Poisson and He lmholtz Propagators in 2D 369 The equations on the left are inhomogeneous partial differential equations driven by source function s( x). If one is careful to include in s( x) all source contributions (such as those on boundary curves), one generally does not have to add any homogeneous solutions on the right. A homogeneous solution refers to -∇2 fh(x) = 0, for example. The solutions shown on the ri ght above can be instantly verified as follows: f(x ) = ∫d2x' [ln(1/R)/2π ] s(x') + fh(x) -∇2 f(x) = ∫d2x' (-∇2 [ln(1/R)/2 π] ) s( x') -∇2 fh(x) = ∫d3x' δ(r-r') s(x') - 0 = s( x) (I.1.10) and similarly for - ( ∇2+k2) f = g. A "free space" Green's Function g F in general is a solution of D gF(r, r') = δ(r-r'), g F(r, r') → 0 as r → ∞ (I.1.11) where D is some differential operator. The condition on the right says g F must vanish on the Great Circle. More generally one can define a full Green's function by, D g(r , r') = δ(r-r'), g( r, r') = 0 for r on some closed curve enclosing a region of interest (I.1.12) This non-free-space Green's function is briefly discussed in the text surrounding (1.5.11). Looking at f( x) = ∫d2x' [ln(1/R)/2π ] s(x') = ∫ gF(x,x') [s( x') d2x'], one can say that the kernel Green's Function g F(x,x') "propagates" a tiny piece of "source" [s( x')d2x'] from location x' to location x so that the solution f( x) is then a sum of all such propagated contributions as the source ranges over the entire area of interest, which for us is all 2D space where the source is non-vanishing. __________________________________________________________________________________ I.2 Derivation of Fact 1: ∇ 2[ln(1/r)] = - 2 πδ(r) ( I . 2 . 1 ) Proof: Let area A be all of 2D space. Cut out from A a small circular hole of radius a centered at r = 0. If we call this circular area A a and then A' = A - A a is the original area with the circular hole cut out: F i g I . 1 In order to show that some function g( r) = δ(r), one has to show that Appendix I : Poisson and He lmholtz Propagators in 2D 370 lima→0 ∫A'dA g( r) = 0 ( I . 2 . 2 a ) lima→0 ∫AadA g( r) = 1 . ( I . 2 . 2 b ) This is basically the definition of δ(r). Since δ(r) has units L-2, g(r) has units L-2. Comment : When any differential operator like ∇ or ∇2 is applied to ln(r 0/r), the result is independent of r0 so we can always take r 0 = 1. For example, ∂ x [ln(r0/r)] = ∂x [ lnr0 + ln(1/r)] = ∂x ln(1/r). In what follows, ln(r) and ln(1/r) are always acted upon by di fferential operators, so we can interpret these objects as dimensionless quantities ln(r/r 0) and ln(r 0/r) for any r 0. Then it is clear below that dim [g(r)] = L-2. Our candidate function of interest is g(r) = - (1/2 π) ∇2[ln(1/r)] = +(1/2 π) ∇2 [ ln(r) ] . (I.2.3) Using ∇ 2 in polar (cylindrical without the z) coordinates acting on a function of r, one finds that, since ∂r(1) = 0, ∇ 2[ln(r)] = (1/r)∂ r(r∂r) [ln(r)] = 0 r > 0 (I.2.4) so that g(r) = - (1/2 π) ∇ 2[ln(1/r)] = 0 r > 0 . (I.2.5) Thus, condition (I.2.2a) is trivially satisfied since r > 0 everywhere in area A' for any a > 0. It remains to verify condition (I.2.2b). Consider the integral appearing in the left side of (I.2.2b) ∫AadA g(r) = - (1/4 π) ∫AadA ∇2[1/r] = - (1/4 π) ∫AadA ∇ • ∇[1/r] . (I.2.6) The divergence theorem (1.1.30) says, in 2D, ∫A dA div F = ∫{C ds • F ( I.2.7) where A is any closed area whose bounding curve is C, and where d s = ds n^ where n^ is normal to C at any given point on C. Notice that this closed area is necessarily planar since everything is 2D here. Using A = A a = disk of radius a and F = ∇[ln(1/r)] = r^∂r(ln(1/r)) = - r^∂r(lnr) = [ -r-1] r^ we find that Appendix I : Poisson and He lmholtz Propagators in 2D 371 LHS (I.2.7) = ∫AadA div ∇[ln(1/r)] = ∫AadA ∇2[ln(1/r)] = ∫AadA [-2πg(r)] = -2π ∫AadA dA g(r) RHS (I.2.7) = ∫C ds • ∇[ ln(1/r)] = ∫ [adθ r^] • ∇[ ln(1/r)]|r=a = ∫dθ [a r^]• [-a-1r^] = -2π which tells us that ∫Aa dA g(r) = 1 for any a. Thus, lim a→0 ∫Aa dA g(r) = 1 and we have then verified (I.2.2b). Ther efore we conclude that the candidate g( r) of (I.2.3) is in fact the same as δ(r) so - (1/2 π)∇2[ln(1/r)] = δ(r) ( I . 2 . 8 ) or ∇ 2[ln(1/r)] = - 2 πδ(r) ( I . 2 . 9 ) w h i c h i s ( I . 2 . 1 ) . QED __________________________________________________________________________________ I.3 Derivation of Fact 2: - ( ∇2+k2) [(j/4) H 0(1)(kr)] = δ (r) (I.3.1) We seek the solution E(r) of this equation - (∇2+k2 ) E(r) = δ (r) where E(r →∞) = 0 (I.3.2) which we write as ∇ 2E+ k2E = - δ(r) . In polar coordinates this says r -1∂r(r∂rE) + k2E = - δ(r) or E" + r -1E' + k2E = - δ(r) or r 2E"(r) + rE'(r) + r2k2E(r) = - δ(r) . ( I . 3 . 3 ) Writing E(r) = F(kr) we get r 2k2F"(kr) + rk F'(kr) + r2k2F(kr) = - δ(r) or (rk)2F"(kr) + (rk) F'(kr) +(rk)2F(kr) = - δ(r) Appendix I : Poisson and He lmholtz Propagators in 2D 372 or z2F"(z) + z F'(z) + z2F(z) = - δ(r) where z ≡ kr . (I.3.4) Away from r = z = 0, this is Bessel's equation of index 0 (NIST 10.2.1) so solutions are Bessel functions like these, F(z) = J 0(z), Y0(z), H0(1)(z), H0(2)(z). z = kr (I.3.5) which are Bessel functions of the first, second and third kind. The third kind functions (the H's) are called Hankel Functions. If we assume that k has a tiny positiv e imaginary part (see Comments later), then of all the functions just listed, only H 0(1)(kr) has decaying behavior for large r (NIST 10.2.5). We therefore put forward the following candidate for a delta function g( r) = - (∇2+k2 ) C H0(1)(kr) . (I.3.6) Recall from Section I.2 that a successful δ(r) candidate must satisfy these two conditions (same as in the previous section, and same figure), lima→0 ∫A'dA g( r) = 0 (I.2.2a) lima→0 ∫AadA g( r) = 1 (I.2.2b) Fig I.1 Our candidate g(r) vanishes within any region A' no matter how small the hole because g(r) = 0 for any r > 0, so the first condition is already met. It rema ins only to show that the second condition is also met. We must then show that lim a→0 ∫Aa dA {- ( ∇2+k2 ) C H0(1)(kr)} = 1 . (I.3.7) Since Aa is a very small disk as we approach the limit, we may use the small argument behavior of our candidate g(r) in studying the situation. We know that H 0(1)(kr) ≈ (2j/π ) ln(kr) // NIST 10.7.2 (I.3.8) so what we need to show is that Appendix I : Poisson and He lmholtz Propagators in 2D 373 lima→0 ∫AadA {- ( ∇2+k2) C (2j/π) ln(kr)} = 1 or - C (2j/ π) lima→0 ∫AadA { ( ∇2+k2) ln(kr)} = 1 or - C (2j/ π)2π lima→0 ∫0 a rdr{ ( ∇2+k2) ln(kr)} = 1 // ∫dθ = 2π or C (4/j) lim a→0 ∫0 a rdr{ ( ∇2+k2) ln(kr)} = 1 . (I.3.9) Now consider : lim a→0 [ ∫0 a rdr ln(kr)] = lim a→0 [(1/4)a2{2ln(ka)-1}] = 0 . (I.3.10) Thus, the k2 ln(kr) term in (I.3.9) makes no contribution in the limit, so we then have to show that C (4/j) lim a→0 ∫0 a rdr ∇2 [ ln(kr)] = 1 . (I.3.11) But (I.2.9) says that ∇ 2[ln(r)] = 2 πδ(r) . (I.2.9) Now δ(r) = δ(x)δ(y) = δ (r)/2πr ( I . 3 . 1 2 ) since 1 = ∫∫dxdy δ(x)δ(y) = ∫rdr∫dθ δ(r)/2πr = 2π∫rdr δ(r)/2πr = ∫dr δ(r) = 1 . Therefore ∇ 2[ln(r)] = δ(r)/r (I.3.13) and then ∇2[ln(kr)] = ∇2[ln(k) + ln(r)] = ∇2[ln(r)] = δ(r)/r . (I.3.14) Inserting this last result into (I.3.11) then gives, Appendix I : Poisson and He lmholtz Propagators in 2D 374 C (4/j) lim a→0 ∫0 a rdr ∇2 [ ln(kr)] = 1 C (4/j) lim a→0 ∫0 a rdr δ(r)/r = 1 C (4/j) lim a→0 ∫0 a dr δ(r) = 1 C (4/j) lim a→0 1 = 1 C (4/j) = 1 . Thus, we have a solution if we select constant C = (j/4). Therefore, the solution to (I.3.2) is E(r) = C H 0(1)(kr) = (j/4) H 0(1)(kr) . (I.3.15) Stakgold Vol II page 55 (5.120) confirms this result where λ = k. Therefore we have shown that - (∇2+k2) [(j/4) H 0(1)(kr)] = δ (r) ( I . 3 . 1 6 ) which is the Fact stated as (I.3.1). QED On page 54 Stakgold gives the solution to - ( ∇2+k2 ) E(r) = δ (r) for n ≥ 2 dimensions as (5.118): Comments: 1. Complex Helmholtz Parameter and H ν(1)(z). Stakgold considers the He lmholtz parameter to be λ which is our k2. He regards λ as a complex variable which can lie anywhere in the complex λ plane. If we consider the function k( λ) = λ1/2, we find that it has a branch point at λ = 0. If we take the branch cut to the right, then one of the two Riemann sheets in λ -space for this function maps to the upper half k-plane as shown. This is the branch of λ1/2 that Stakgold selects and that is why we think of k and therefore k2 as having a tiny positive imaginary part when k is "real". The point is that we approach the positive real axis from above, not from below. It is this assu mption that causes the large-r-decaying solution to our problem to be H 0(1)(kr) instead of H 0(2)(kr) . Fig I.2 Appendix I : Poisson and He lmholtz Propagators in 2D 375 As shown on NIST p 229 10.17.5,6, expansions of the Hankel functions for large argument are, Hν(1)(z) ≈ 2/π z-1/2 e+j(z-νπ/2-π/4) Σk=0∞ (+j)k ak(ν) z-k Hν(2)(z) ≈ 2/π z-1/2 e-j(z-νπ/2-π/4) Σk=0∞ (-j)k ak(ν) z-k where a k(ν) are some real coefficients shown in 10. 17.1 which we don't care about right now. The differences are highlighted in red. Here one sees that H ν(1)(z) ~ e+jz = e-Imz ejRez. Thus H ν(1)(kr) ~ e-rImk ejrRek and as long as k is in the upper half plane as shown in the right, H ν(1)(kr) decays exponentially (whereas H ν(2)(kr) blows up). 2. Helmholtz morphs into Poisson. We have shown that - (∇2+k2) [(j/4) H 0(1)(kr)] = δ (r) . (I.3.16) In the limit that k << 1, we showed above that H0(1)(kr) ≈ (2j/π ) ln(kr) // A&S 10.7.2 (I.3.8) In this limit we then have - (∇ 2+k2) [(j/4)) (2j/ π) ln(kr) ] = δ(r) or - (∇ 2) [(1/2π) ln(kr) ] = δ(r) and this is in agreement with the Poisson result (I. 1.1). So as the Helmholtz equation morphs into the Poisson equation as k → 0, the Helmholtz propagator morphs into the Poisson propagator. Appendix J : The 3D→2D Propagator Transition 376 Appendix J: The 3D →2D Propagator Transition Infinitely long transmission lines ar e -- in the transmission line limit of long wavelength -- basically 2D objects rather than 3D objects. We see that fact ap pearing in various Chapters and Appendices of this document. Here we wish to focus on this single fact. Case 1 In Chapter 1 we presented the natural 3D view of transmission lines with equations like the following taken from (1.5.3), (1.5.4) and (1.5.23), where we used the King gauge, ( ∇ 2 + βd2)φ = - (1/ε) Σiρi ⇔ φ(x,ω) = 1 4πξd Σi∫ρci(x',ω) e-jβdR R dV' (∇2 + βd2)A = - ΣiμiJi ⇔ A (x,ω) = 1 4π Σi∫μiJi(x',ω) e-jβdR R dV' ., (J.1) The Helmholtz integrals on the right are particular solutions of the PDEs on the left. The equations on the right are derived from those on the left as shown in Appendix H where we had the more generic statement that - (∇ 2+k2) f(x) = s( x) => f( x) = ∫d3x' [e-jkR/4πR] s( x') + homogeneous solutions The 3D Helmholtz Equation particular solution (H.1.9) The object [e-jkR/4πR] is the 3D free-space Helmholtz propagator as discussed in Appendix H. If it happens that f( x) = f(x,y) in this last equation, then ∂ z2f = 0 and we find ourselves looking at a 2D Helmholtz equation which has a completely di fferent-looking particular solution, where ∇2 = ∇2D2 + ∂z2, - (∇2D2+k2) f(x) = s( x) => f( x) = ∫d2x' [(j/4) H 0(1)(kR)] s( x') + homogeneous solutions The 2D Helmholtz Equation particular solution (I.1.9 ) This is the most abrupt and simple way the transition from 3D to 2D can occur. If k is small, meaning the corresponding wavelength λ = 2π/k is large, we can take the small k limit of the above two particular integrals. The limit of [e -jkR/4πR] is completely obvious, [e-jkR/4πR] → [1/4πR ] ( J . 2 ) whereas the limit of the 2D propagator [(j/4) H 0(1)(kR)] is less obvious: H 0(1)(kr) ≈ (2j/π ) ln(kr) // NIST 10.7.2 (I.3.8) Appendix J : The 3D→2D Propagator Transition 377 so that [(j/4) H 0(1)(kR)] ≈ - (1/2π) ln(kR) = [- (1/2 π) ln(R)] - (1/2 π) ln(k) . (J.3) If we momentarily ignore the inconvenient constant - (1/2 π) ln(k), we can say that 2D Helmholtz propagator = [j 4 H0(1)(kR)] → [- 1 2π ln(R)] = [-1 4π ln(R2)] = [1 2π ln(1/R) ] . (J.4) The objects on the right of (J.2) and (J.4) and are in f act the 2D Poisson propagators which belong to this pair of PDE's and their particular solutions, -∇ 2 f(x) = s( x) => f( x) = ∫d3x' [1/4πR] s( x') + homogeneous solutions The 3D Poisson Equation particular solution (H.1.8) -∇2D2 f(x) = s( x) => f( x) = ∫d2x' [ln(1/R)/2π ] s(x') + homogeneous solutions The 2D Poisson Equation particular solution (I.1.8) Since in our applications f( x) is always a potential like φ or A , and since B = curl A E = - grad φ - ∂tA (1.3.1) we see that a constant like - (1/2 π) ln(k) added to a potential has no effect on the physical fields E and B, so we can just ignore such constants. Another way to say this is that the zero level of a potential is always arbitrary so additive constants are meaningless. In Chapter 4 we are only really concerned with the potential difference V(z) or W(z) between conductors. We can now look at some of the 3D/2D "transitions" that occurred in other parts of the document. Case 2 In Section 4.4 we had V(z) ≡ φ 12(x1) - φ12(x2) = 1 4πξd q(z) ∫-∞ ∞ dz'{ ∫C1 dx1' dy1' α1(x1',y1') 1 R11 – ∫C2 dx2' dy2' α2(x2',y2') 1 R12 } – 1 4πξd q(z) ∫-∞ ∞ dz'{ ∫C1 dx1' dy1' α1(x1',y1') 1 R21 – ∫C2 dx2' dy2' α2(x2',y2') 1 R22 } (4.4.1) which we obtained by assuming a separated form (4.1.2) for the charge density and by assuming a small Helmholtz parameter β. The 1/4 πR factors here are in fact the 3D Poisson free-space propagators. This propagator has the less glamorous name of being the electrostatic potential of a (1/ ε)-size point charge (in "free space" of course), so by assuming the transmission line limit of small Helmholtz parameter β , we Appendix J : The 3D→2D Propagator Transition 378 arrive at this electrostatics Poisson propagator appearing in the integrals. These propagators are "propagating" the effect of charges on th e conductor surfaces to their destinations x1 and x2 in Fig 4.2 . We then did the dz' integral over (- ∞,∞) making use of integral (4.4.5), ∫-∞ ∞ dz' ( 1 R12 - 1 R22 ) = ln(s 222/s122) (4.4.5) and arrived at V(z) = q(z) 1 4πξd { ∫C1 dx1' dy1' α1(x1',y1') ln(s212/s112) - ∫C2 dx2' dy2' α2(x2',y2') ln(s222/s122) } . (4.4.6) This is really four terms and one recognizes -1 4π ln(R2) in the form -1 4π ln(sij2) as the 2D Poisson propagator just discussed above, and the s ij are the 2D transverse distances shown in Fig 4.3. So here we see a very clear example of doing the 3D → 2D transition. Case 3 Another transition example is the "scaling boundary cond ition" of Section 5.3 (b). We started there with φ t(x) = ∫-∞ ∞ dz'{ ∫C1 dx1' dy1' α1(x1',y1') 1 R1 – ∫C2 dx2' dy2' α2(x2',y2') 1 R2 } (5.1.2) and we moved the observation point x far away from the transmission lin e. The result in this limit was found to be φ t(x) ≈ ln(s22/s12) // limiting form as point x = (x,y) moves far from the conductors (5.3.13) In this case, we had earlier done the following separation of the full potential φ(x,y,z) = 1 4πεd q(z) φt( x , y ) (5.1.1) so the limit shown for φt says φ(x,y,z) ≈ (q/εd) 1 4π ln(s22/s12) = (q/εd) 1 4π ln(s22) – (q/εd) 1 4π ln(s12) (J.5) and we interpret this as being the sum of the 2D free-space propagations of charges ±q(z)dz to our distant point. We are so far from the transmission line that these charges appear as 2D point charges which form a little electric dipole as shown in Section 5.4 (b). Appendix J : The 3D→2D Propagator Transition 379 Case 4 As a third example, we consider a simple generic situation alluded to above as our "abrupt" transition. Start again with - (∇ 2+k2) f(x) = h(x ) => f( x) = ∫d3x' [e-jkR/4πR] h( x') + homogeneous solutions The Helmholtz Equation particular solution (H.1.9) We changed the source name from s(x) to h(x) to avoid confusion with distance s below. We now assume that f( x) = f(x,y). What happens to the Helmholtz integral on the right? f(x ) = ∫d3x' [e-jkR/4πR] h( x') = ∫dx' ∫dy' ∫-∞ ∞ dz' h(x',y') [e-jkR/4πR] where s = (x-x')2 + (y-y')2 and R = s2 + z'2 . Then f(x ) = ∫dx' ∫dy' h(x',y') 1 4π ∫-∞ ∞ dz' e-jkR R R = s2 + z'2 . We can do the dz' integral as follows: R 2 = s2+ z'2 => RdR = z'dz' so ∫-∞ ∞ dz' e-jkR R = ∫-∞ ∞ RdR z' e-jkR R = ∫-∞ ∞ dR e-jkR R2-s2 = ∫-∞ ∞ dR cos(kR) R2-s2 = 2 ∫0 ∞ dR cos(kR) R2-s2 . ( J . 6 ) We then take note of the following integral in GR7 3.754.2 page 435, which then says ∫-∞ ∞ dR cos(kR) R2-s2 = K0(k-s2 ) = K0(-jks) z = -jks phase (z) = - π/2 (J.7) ( z eπj/2) = zj = ks But NIST p 250 says Appendix J : The 3D→2D Propagator Transition 380 so that K 0(-jks) = π(j/2)H0(1)( k s ) ( J . 8 ) and then ∫-∞ ∞ dz' e-jkR R = 2 ∫-∞ ∞ dR cos(kR) R2-s2 = jπH0(1)(ks) . (J.9) Finally f(x ) = ∫dx' ∫dy' h(x',y') 1 4π ∫-∞ ∞ dz' e-jkR R R = s2 + z'2 = ∫dx' ∫dy' h(x',y') 1 4π jπH0(1)(ks) = ∫dx' ∫dy' h(x',y')[ (j/4) H 0(1)(ks)] s2 (x-x')2 + (y-y')2 and once again we have transitioned from the 3D propagator e-jkR R to the 2D one (j/4) H 0(1)(ks). Case 5 In the k = 0 limit the above Case becomes a transition from 3D propagator 1 R to 2D propagator -1 2π ln(s) as follows : f(x ) = ∫d3x' [1/4πR] h( x') = (1/4 π) ∫dx' ∫dy' h(x',y') ∫-∞ ∞ dz' s2 + z'2 . But now the dz' integral is logarithmically divergent so we install a very large cutoff Λ and write ∫-∞ ∞ dz' s2 + z'2 → ∫-Λ/2 Λ/2 dz' s2 + z'2 = 2 ∫0 Λ/2 dz' s2 + z'2 = 2 ln[ z' + z'2 + s2 ] | Λ/2 0 = 2ln[ Λ/2 + (Λ/2)2 + s2 ] - 2 ln s Appendix J : The 3D→2D Propagator Transition 381 ≈ 2ln(Λ) - 2lns = -2ln(s/ Λ) . ( J . 1 0 ) Now we apply the argument above about ignoring constants to get, ∫-∞ ∞ dz' s2 + z'2 = - 2lns // ignoring constants f(x ) = ∫dx' ∫dy' h(x',y') (1/4 π) (-2lns) = ∫dx' ∫dy' h(x',y') [-1 2π ln(s)] and so we have transitioned in this case from the 3D Poisson propagator to the 2D one. Appendix K : The Network Model 382 Appendix K: The Network Model: Comparison of Network a nd Maxwell Views (a) The Network Model The usual net work model of a 2-conductor transmission li ne is an infinite repetition of differentially small R,L,C,G segments as shown here between the vertical red lines, Fig K.1 If some load Z L is put on the right end of Fig K.1, the im pedance seen from the left end is unchanged if the segment circuit is replaced by the following equivalent circuit, Fig K.2 where R = R 1 + R2 and L = L 1 + L2. In the circuit diagrams, it is implied that R,L,C,G are all quantities per unit length of the transmission line. Thus, if the distance between the two red lines is δ , the values of the lumped parameters in Fig K.2 are R δ,Lδ,Cδ,Gδ. For example, if δ doubles, the total conductance of the segment doubles since it is a measu re of current flowing between the conductors. The model implied by the picture is then the limit as δ→0. We wish to compare this "network model" to our Maxwell equation results. To start, we note that the impedance of a capacitor C and in ductor L operating at frequency ω are determined by Q = CV => I = ∂ tQ = C ∂tV => I = j ωCV => Z C = V/I = 1/(j ωC) V = L ∂ tI => V = j ωLI => Z L = V/I = jω L . (K.1) Note that the "admittance" of a capacitor is Y C = 1/ZC = jωC. We can then combine the G and C elements together into a single element having y = G+jωC, since parallel admittances are additive. Similarly, we combine the two series elements R and L into impedance z = R+jωL. The network picture is then, Appendix K : The Network Model 383 Fig K.3 We use King's bolded symbols y and z and of course z is unrelated to distance z. Here we arbitrarily have the z axis pointing to the left (!), and the vertical red lines are placed at z and z+dz so that δ = dz. The impedance looking into the transmission line from the left is Z(z+dz) at z+dz and is Z(z) at z. (b) Network Model Characteristic Impedance Since the im pedance 1/ y is in parallel with the impedance z + Z(z) one has Z(z+dz) = (y dz)-1 || (zdz + Z(z)) = product over sum = (ydz)-1 (zdz + Z(z)) (ydz)-1 + (zdz + Z(z)) = (zdz + Z(z)) 1 + ( ydz) (z dz + Z(z)) ≈ [Z(z) + zdz] [1 - ( ydz)(z dz + Z(z)] ≈ [Z(z) + zdz] [1 - ( ydz) Z(z)] // dropping order (dz)2 ≈ Z(z) + [ z - yZ 2(z)] dz . // dropping order (dz)2 again Therefore Z(z) must solve this non-linear first order differential equation, dZ(z) dz = z - yZ2(z) or dZ(z) dz + y Z2(z) = z . (K.2) The most general solution to this equation is Z(z) = z y th ( zy z + C ) C = constant (K.3) since ∂ zZ = z y * zy sech2(z y z + C ) = z [ 1 - th2(z y z + C ) ] = z [ 1 - y z Z2(z)] = z - y Z2(z) . If the transmission line is of finite length running from z = L (left end ) to z = 0 (right end), and if the line is terminated at z = 0 by some impedance Z t, we must have Z(0) = Z t so that Appendix K : The Network Model 384 Zt = Z(0) = z y th ( zy 0 + C ) = z y th(C) => C = th-1(y z Zt) so then the solution is Z(z) = z y th [ zy z + th-1(y z Zt) ] and at the left end we find Z(L) = z y th [ zy L + th-1(y z Zt) ] . If we take L → ∞ (line becomes infinitely long) , then th [...] → 1 and we find Z(∞) = z y , so the impedance looking into the left end of the infinite transmission line is independent of the termination value Z t at z = 0. This infinite line impedance is called the characteristic impedance Z0 and we have shown then that Z 0 = z y = R+jωL G+jωC . ( K . 4 ) Since this is the same result obtained from Maxwell's equations in (4.11.16), one is motivated to regard the network transmission line model as a correct model, and then the netw ork model parameters R,L,G,C can be identified with the parameters obtained from Maxwell's equations. Reader Exercise: Consider this purely resistive finite la dder network shorted at the right end, Fig K.4 Appendix K : The Network Model 385 (1) Using the results above, show that R(L) = R3/G tanh ( R3G L ) where R 3 ≡ R1+ R2 . (2) Show that R(L) ≈ R3/G if L >> 1/ R3G . Thus, for large L the fact that the line is shorted at the right end makes no difference. (3) Show that for finite L : R(L) → R 3L as G → 0 no conductance R(L) → 0 as R 3→ 0 no wire resistance Both limits should seem obvious. (c) Network Model Transmission Line Equations We now swit ch the z axis back to its usual directi on (increasing to the right), and we label currents and voltages on our transmission line section, Fig K.5 Staring at the picture, it seems clear that i(z) - i(z+dz) = current going down through impedance 1/( ydz) = V(z) 1/(ydz) = ydz V(z) and therefore - di(z) dz = y V(z) . Meanwhile, the voltage across the impedance z is V(z) - V(z+dz) so V(z) - V(z+dz) = i(z) z and therefore Appendix K : The Network Model 386 - dV(z) dz = z i(z) . Thus we have shown that dV(z) dz = - z i(z) di(z) dz = - y V(z) with z = R + jωL y = G +jω C . ( K . 5 ) Differentiating these equations with respect to z, we find that d 2V(z) dz2 - zy V(z) = 0 d2i(z) dz2 - zy i(z) = 0 (K.6) But (K.5) and (K.6) are the same transmission line equations obtained from Maxwell's equations as shown in (4.11.14b) and (4.11.15). Thus we are furt her encouraged in our use of the network model to represent a transmission line. Since the equations fo und from Maxwell's equations were qualified as being questionable at very low frequencies, the network model is also suspect at very low ω (d) Network Model Parameters obta ined from M axwell's Equations The main results of Chapter 4 appear in summary box (4.11.34) from which we quote in part, dV(z) dz = - z i(z) ( d2 dz2 - zy) V(z) = 0 z = R + jωL transmission line equations di(z) dz = - yV(z) (d2 dz2 - zy) i(z) = 0 y = G +jω C (4.11.14) , (4.11.15) z = Zs1 + Zs2 + jωLe (4.11.17) XL ≡ ωLe , XC ≡ 1/(ωC) y = jω C' = jωC + (σd/εd)C (4.11.24) G = (σ d/εd)C (4.11.25) R = Re(Z s1+ Zs2) L = L e + (1/ω) Im(Zs1+ Zs2) Le = (μd/4π)K (4.11.29) and (4.11.30) C = 4πεd/K (4.11.26) G = 4πσd/K (4.11.26) + (4.11.25) ( K . 7 ) Thus, we make the connection betw een the network parameters and th e Maxwell calculation parameters as follows: R = Re(Z s1+ Zs2) L = L e + (1/ω) Im(Zs1+ Zs2) Appendix K : The Network Model 387 Le = (μd/4π)K G = 4 πσd/K C = 4 πεd/ K ( K . 8 ) where K is the dimensionless real integral in Chapte r 4, see (4.4.8). Recall that this integral requires knowledge of both the conductor geometry as well as the normalized transverse surface charge distributions on the conductors. Here ε d, μd and σd are for the dielectric between the conductors. The effective σd appearing in G is often frequency dependent as shown in (3.3.4) [see Appendix R for an example]. The Z si are always frequency dependent, as discussed below. (e) Low frequency case (no skin effect) At low frequencies, when conductors are not extremel y close together, the current densities are close to uniform (see for example Fig 6.16), so that J z = I/area for each conductor. This uniformity is exact for a conductor which is the central conductor of a coaxial cable, as studied in Chapter 2. There we found at low frequency that Z s1(ω) = 1 σ1πa2 + jω μ1 8π // low frequency limit (2.4.12) => Re(Z s1) = 1 σ1πa2 and Im(Z s1) = ω μ1 8π . (K.9) From (K.8) we then find that for low frequencies and paralle l round conductors, R = 1 σ1πa12 + 1 σ2πa22 = Rdc1 + Rdc2 ( K . 1 0 ) L = L e + ( μ1 8π + μ2 8π ) = Le + (Li1 + Li2) . ( K . 1 1 ) In this case parameter R is just the sum of the DC resistances of the conductors (per unit length), and parameter L is the sum of the external inductance L e and the internal inductances of the two wires. Here σi and μi are for the material from which conductor C i is constructed. The external inductance L e can be interpreted as the inductance of the red wire loop below, Fig K.6 Appendix K : The Network Model 388 The sides of the red loop make contact on any lin e on the conductor surfaces, though here we show it having its minimal size. The red loop may in fact be replaced by any loop, possibly non-planar, which captures all the external magnetic flux passing between the conductors. See Fig 4.11 and discussion there. Note that L e is not the self-inductance of a rectangular thin wire loop in isolation occupying the red outline above, but rather L e = (μd/4π)K as in (K.8) above, where K is related to the capacitance between the conductors. If both conductors are round and very thin and separated by distance b, we know from (4.5.7) that K = 4 ln(b/ a1a2 ) and then L e = (μd/π) ln(b/ a1a2 ). Although we have not formally proven it, it seems clear that for arbitrary conductor cross sections (not too closely spaced) the following equati ons will apply at low frequency : R = 1 σ1A1 + 1 σ2A2 A i = cross section area of C i (K.12) L = L e + (Li1 + Li2) . ( K . 1 3 ) Appendix C computes the DC L i for various conductor cross section shapes. One result quoted there from the literature is that for a square conductor, Li = (μi/8π) [0.96639] . (C.4.12) Thus the L i for a square cross-section conductor is barely different from that of a round conductor. (f) High frequency case for round conductor (strong skin effect) At high frequencies there is a pronounced skin effect. In Chapter 2 for a ro und conductor C 1 at high frequency (and with a symmetric current distribution) we found that Zs1(ω) ≈ 1 σ1(2πa1)δ1 (1+j) δ1 << 4a1 (2.4.16) so Re(Z s1) = Im(Z s1 ) = 1 σ1(2πa1)δ1 = 1 2πa1 μ1ω 2σ1 where δ1 = 2/(ωμ1σ1) is the skin depth and a 1 the wire radius. From (K.8) we find that for high frequencies and round conductors, R = 1 σ 1(2πa1)δ1 + 1 σ2(2πa2)δ2 L = L e + (1/ω) Im(Zs1+ Zs2) = Le + (1/ω) R . ( K . 1 4 ) Appendix K : The Network Model 389 In this case, we recognize 2 πa1δ1 as the effective current carrying cross-sectional area of round conductor C1 (the area of the current sheath), so the expression for R is quite intuitive. Since, δ ≡ 2/ωμσ => 1/ δ1 = (ωμ1σ1)/2 and 1/ ω = μ1σ1δ12/2 (K.15) we may write Li(ω) = (1/ω) Im(Zs) = (1/ω) 1 σ1(2πa1)δ1 = (1/ω) 1 σ1(2πa1) (ωμ1σ1)/2 = 1 2πa1 μ1 2σ1ω ( K . 1 6 ) so Li(ω) ~ 1/ω . Expressing L i instead in terms of δ1 we find Li(δ1) = (1/ω) Im(Zs) = (1/ω) 1 σ1(2πa1)δ1 = μ1σ1(δ12/2) 1 σ1(2πa1)δ1 = μ1 (1/4π) (δ1/a1) = μi 8π [ 2 (δ1/a1) ] . ( K . 1 7 ) The DC internal inductance of a thin shell of radius a and thickness d is show n in Appendix C.6 to be Li = μi 1 6π (d/a) = μi 8π [ (4/3)(d/a) ] thin shell, valid for d << a (C.6.8) so the high frequency internal inductance of a round wire is the same as the DC internal inductance a shell of thickness d = (3/2) δ which seems fairly reasonable. The a bove expression (C.6.8) shows that the inductance of a thin cylindrical shell is linear in the shell thickness d, so we expect that the high frequency L i of a round wire should be linear in δ, and thus proportional to 1/ ω . Section 2.5 shows how to handle non-round conduc tors and non-symmetric current distributions by replacing 2 πa by an effective active perimeter p. Chapte r 4.11 (b) formalizes this notion, giving the effective perimeter in (4.11.10). In Section D.10 and D.11 the claims made in th e last two sections regarding surface impedance are vindicated when one uses the surface impedance averaged over the round wire surface, see (D.10.17) for large ω and (D.11.10) for low ω. Appendix L : Point and Line Charges in Dielectrics 390 Appendix L: Point and Line Charges in Dielectrics Chapter 1 states in (1.1.19) throug h (1.1.24) various equations concerni ng the magnetization of a magnetic medium. These equation are "exercised" so mewhat in Section G.3 and also in Appendix B concerning how the transmission line theory is altere d when the dielectric and conductors have different μ values. Chapter 1 also states in (1.1.9) through (1.1.15) co rresponding equations concerning the polarization of a dielectric medium. Although the transmission line th eory assumes a dielectric between the conductors, and in fact allows for a complex dielectric constant ξ, there has been no "exercise" of the polarization equations, so in this Appendix so me simple examples are provided. The examples presented here are rarely presented in E&M texts perhaps because they are too simple. The spherical problem appears in the 2nd edition of Corson and Lorrain (p 111-113) but it got replaced by a short comment in the 3rd edition (Corson and two Lorrains) p 186. The examples are useful to the author in that they provide a physical picture of how the potential and field of a point or line charge are affected by the presence of a dielectric medium. In Sections L.1 and L.2 the 3D problem is solved and limits are taken of the solution. Two of these limits involve a full embedding of the charge in the dielectric where dielectric charge shielding is exhibited. S ections L.3 and L.4 briefly repeat the solution in two dimensions, so the r esults then apply to the extruded cross section. L.1 The potential of a point charge inside a thick dielectric spherical shell. A positive point charge q li es at the center of a spherical shell of radii b>a as follows, Fig L.1 Inside and outside the shell of dielectric constant ε1 is empty space with ε0. Whatever the potential φ is for the above picture, it is obviously spherically symmetric and is then φ(r). This in turn means that the E field is just E = Err^ where E r = -∂rφ (in each region), so the E field is Appendix L : Point and Line Charges in Dielectrics 391 radial. This radial E field polarizes the dielectric in the shell as suggested by the three symbolic polarized molecules shown in the figure. If the total bound charge on the r = a surface is -Q, then the total charge on the r = b surface must be +Q, as one would conclude imagining the entire dielectric having the form of the three molecules shown. One implication of this fact is that for a sphere of r > b, the total charge enclosed is just q. Applying Gauss's law to a spherical Gaussian box of radius r > b q = ∫V ρ dV = ∫S ε E • dS = ε0[∫dΩ] r2 E • drr^ = ε04π r2Er (1.1.33) => E r0 = ( 1/4πε0) q/r2 . The corresponding potential is φ 0(r) = (1/4πε0) q/r r > b region 0 since then E r0 = -∂rφ0(r) = (1/4πε0)q/r2. This is a special case of the fact that any spherical distribution of charge appears outside that distribution as a point charge at the center, so φ 0(r) is just the potential of a point charge q at the origin. We then at least know φ in one of the three regions. In regions 1 and 2 as an ansatz we assume these forms with constants α,C and D to be determined, φ 1(r) = (1/4πα ) q/r + C φ2(r) = (1/4πε0) q/r + D . In a spherically symmetric geometry the Laplace equation only allows harmonics that are powers r n and each term above is one such power times a constant. A motivation for the φ 2 form is that for r very close to r = 0, the potential must be that of the point ch arge since everything else is then relatively far away. We now determine constants B,C and D from boundary conditions. The three potentials and fields are φ0(r) = (1/4πε0) q/r E r0(r) = (1/4πε0) q/r2 region 0 φ1(r) = (1/4πα ) q/r + C E r1(r) = (1/4πα ) q/r2 region 1 φ2(r) = (1/4πε0) q/r + D E r2(r) = (1/4πε0) q/r2 region 2 . (L.1.1) The electrostatic potential must be continuous at all values of r. Why? Consider: E r = -∂rφ ∫a b Erdr = - ∫a b ∂rφ dr = - [ φ (b) - φ(a) ] . The physical electric field at any point must have a well-defined finite single value. Then for small ε ∫a a+ε Erdr = Er(a) ε = - [ φ(a+ε) - φ(a) ] => φ continuous at a . (L.1.2) As ε → 0, we must have φ(a+ε) → φ(a) so φ(r) must be continuous at r = a. Appendix L : Point and Line Charges in Dielectrics 392 Apply this rule at our two boundaries to find that, (1/4πε0) q/b = (1/4 πα ) q/b + C region 0/1 boundary, r = b (1/4πε0) q/a + D = (1/4 πα ) q/a + C region 2/1 boundary, r = a (L.1.3) which is two conditions on the unknown constants α,C,D. Meanwhile, the normal electric field boundary condition from Chapter 1 is [ε1En1 - ε2En2] = nfree . (1.1.47) Although there exists bound charge at each of our two boundaries, there is no free charge, so ε0Er0(b) = ε1Er1(b) region 0/1 boundary, r = b ε0Er2(a) = ε1Er1(a) region 2/1 boundary, r = a or ε 0 q/ [4πε0b2] = ε1 (1/4πα ) q/b2 => 1 = ε1/α ε0 q/ [4πε0a2] = ε1 (1/4πα ) q/a2 => 1 = ε1/α . (L.1.4) The right side equations are the same and tell us that α = ε 0. The boundary conditions (L.1.3) then say, (1/4πε 0) q/b = (1/4 πε1) q/b + C region 0/1 boundary, r = b (1/4πε0) q/a + D = (1/4 πε1) q/a + C region 2/1 boundary, r = a . (L.1.5) Subtract the first from the second to cancel the C, D + (1/4 πε0) q(1/a-1/b) = (1/4 πε1)q (1/a-1/b) so D = (1/a-1/b)(q/4 π)(1/ε 1-1/ε0) = - (b/a-1)(q/4 πb)(1/ε0-1/ε1) . From the first of (L.1.5) we find C = (q/4 πb) (1/ε 0-1/ε1) . Thus the boundary conditions have determined our three constants α = ε0 C = (q/4πb) (1/ε0-1/ε1) D = (q/4 πb)(1/ε0-1/ε1) (1-b/a) . (L.1.6) Appendix L : Point and Line Charges in Dielectrics 393 The potentials in the three regions are then φ0(r) = (1/4πε0) q/r φ1(r) = (1/4πε1) q/r + (q/4 πb)(1/ε0-1/ε1) φ2(r) = (1/4πε0) q/r + (q/4 πb)(1/ε0-1/ε1) (1-b/a) (L.1.7) while the fields are E r0 = (1/4πε0) q/r2 Er1 = (1/4πε1) q/r2 Er2 = (1/4πε0) q/r2 . ( L . 1 . 8 ) We know that a spherical shell of charge has no effect on E r inside the shell, verifying the E r2 result. We also know that a spherical shell acts as a point charge at the origin when viewed from outside the shell, thus verifying E r0 which then sees a charge of q + Q - Q = q at the origin. The following Maple plots show the continuity of φ and the jumps in E r at the boundaries Fig L.2 φ (r) [red] and E r(r) [blue] for r in (0.5, 5) What about the bound charge densities at r = a and r = b? Appendix L : Point and Line Charges in Dielectrics 394 One must first compute polarization P, and for a region with ε, P is given by P = ε0χeE // polarization assumed proportional to the polarizing E field (1.1.12) so P = ε0χeErr^ χe = (ε/ε0 - 1) => P = ε0(ε/ε0 - 1)Err^ = (ε-ε0)Err^ . Obviously P = 0 in regions 0 and 2, while in region 1 we have Pr1(r) = (ε1-ε0)Er1(r) θ(r>a)θ(r<b) // points radially outward since ε 1 > ε0 (L.1.9) From (1.1.11) the polarization charge density is then ρ pol = - div P (1.1.11) so in spherical coordinates, ρ pol(r) = - [r-2∂r(r2Pr) + [rsinθ]-1∂θ[sinθPθ] + [rsinθ]-1∂φPφ] = - r-2∂r(r2Pr1) = - r -2∂r(r2[(ε1-ε0)Er1(r) θ(r>a)θ(r<b)]) // from (L.1.9) = - r -2∂r( [(ε1-ε0)(1/4πε1) q θ(r>a)θ(r<b)] ) // from (L.1.8) = - q( ε 1-ε0)(1/4πε1) r-2∂r( [θ(r>a)θ(r<b)] ) But ∂r[θ(r>a)θ(r<b)] = ∂r[θ(r-a)θ(b-r)] = δ (r-a) θ (b-r) + θ(r-a) [-δ(r-b)] = δ(r-a) θ(b-a) - θ (b-a) δ(r-b) = [ δ(r-a) - δ(r-b)] , so ρpol(r) = - q( ε1-ε0)(1/4πε1) r-2 [δ(r-a) - δ(r-b)] = - q( ε 1-ε0) (1/4πε1) { δ(r-a)/a2 - δ(r-b)/b2} . (L.1.10) We may then read off the bound surface charge densities at r = a and b, σinner = - (ε1-ε0) (1/4πε1)(q/a2) σouter = (ε1-ε0) (1/4πε1)(q/b2) ( L . 1 . 1 1 ) so Qinner = ∫dS σinner = - (ε1-ε0) (1/4πε1)(q/a2) * 4πa2 = - (1-ε 0/ε1) q Qouter = ∫dS σouter = (ε1-ε0) (1/4πε1)(q/b2) * 4πb2 = (1-ε0/ε1) q . (L.1.12) Appendix L : Point and Line Charges in Dielectrics 395 Thus the outer boundary has total charge Q = (1- ε 0/ε1) q // ranges from 0 to q (L.1.13) and the inner boundary has -Q. If the dielect ric were a conductor, we would replace ε1 → ξ1 as in (1.5.1c) and then a perfect conductor has ξ1 = ∞ and so Q = q, as one would expect looking at Fig L.1. L.2 Limits of the Previous Problem (a) Point charge in a spherical cavity in a dielectric Taking b→∞ in the previ ous problem removes outer region 0 and leaves us with this picture of a point charge at the center of a spherical hole in an infinite medium of ε1 : Fig L.3 The potentials and fields shown in (L.1.7) and (L.1.8) are then, taking b →∞, φ1(r) = (1/4πε1) q/r φ2(r) = (1/4πε0) q/r - (q/4 πa)(1/ε0-1/ε1) (L.2.1) while the fields are Er1 = (1/4πε1) q/r2 Er2 = (1/4πε0) q/r2 . ( L . 2 . 2 ) The induced bound charge density σ at r = a, and the total charge there, are still given by σ = - (ε1-ε0) (1/4πε1)(q/a2) -Q = (1- ε0/ε1) q . ( L . 2 . 3 ) Appendix L : Point and Line Charges in Dielectrics 396 (b) Point charge embedded in a dielectric sphere Here we take the lim it a→0 so that region 2 of Fig L.1 goes away. Looking at (L.1.12), the total inner surface bound charge continues to be - (1- ε0/ε1) q = -Q in this limit. It just crowds around the point charge and of course the surface density σinner → ∞. Here is a suggestive drawing of a piece of region 1 in this limit: Fig L.4 The limiting picture of Fig L.1 is then the following, Fig L.5 The potentials and fields shown in (L.1.7) and (L.1.8) are then, taking a → 0, φ0(r) = (1/4πε0) q/r φ1(r) = (1/4πε1) q/r + (q/4 πb)(1/ε0-1/ε1) (L.2.4) while the fields are E r0 = (1/4πε0) q/r2 Er1 = (1/4πε1) q/r2 . (L.2.5) Appendix L : Point and Line Charges in Dielectrics 397 Inside the dielectric the E field is E r1 = (1/4πε1) q/r2 where ε1 takes into account both the point charge q and the bound charge crowding around it which is - (1- ε0/ε1) q. One could interpret this as saying that the total charge at the origin is q - (1- ε0/ε1) q = q( ε0/ε1) and then E = (1/4 πε0) [q(ε0/ε1)]/r2. Remember from (1.1.15) that E sees both free and bound charge. In th is last interpretation, the dielectric is shielding the point charge, reducing it from q to q( ε0/ε1). Outside the sphere, the E field is E r1 = (1/4πε 0) q/r2, just as if the sphere were not there. The reason of course is that the surface charge at r = b still cancels the crowded surface charge at r = 0, so outside one sees in effect just the point charge q. (c) Point charge embedded in an infinite dielectric medium We now take b →∞ in Fig L.5 to remove outer region 0, with this result: Fig L.6 There is only one region left and from (L.2.4) and (L.2.5) we get φ1(r) = (1/4πε1) q / r ( L . 2 . 6 ) while the field is E r1 = (1/4πε1) q/r2 . ( L . 2 . 7 ) The presence of the dielectric ε 1 is then completely accounted for by the (1/4 πε1) factor. As before, one could interpret this as a shielded charge [q( ε0/ε1)] and E = (1/4 πε0) [q(ε0/ε1)]/r2. The crowded-around polarization charge is still -Q = - (1- ε0/ε1) q, and the positive Q that was on the r = b surface is still present, but at r = ∞. Appendix L : Point and Line Charges in Dielectrics 398 L.3 The potential of a line charge inside a thick dielectric cylindrical shell In this section, we repeat every thing done in Section L.1 in the 2D world instead of the 3D world. The 3D Poisson propagator (1/4 πε0)(1/r) becomes (1/2 πε0) ln(1/r) as discussed in Appendix J. We reuse the same drawings, the first of which is Fig L.1' This is now a cross section of an infinite uniform cylindrical hollow dielectric pipe. Quantity q is now a linear charge density with dimensions Coulombs/m. Rather that copy, paste and edit Section L.1, here we just show the altered equations and skip most of the words. The equation numbers are those of Section L.1 with a prime added. One difference encountered is that we must take b→R (a large value) rather than b→∞. As noted in Appendix J, a constant in a potential can be ignored even if it is infinite, and such constants do not appear in the field E = -∇φ. Ansatz potential forms: (to-be-determined constants are α, C, D) φ 0(r) = (1/2πε0) q ln(1/r) E r0(r) = (1/2πε0) q/r region 0 φ1(r) = (1/2πα ) q ln(1/r) + C E r1(r) = (1/2πα ) q/r region 1 φ2(r) = (1/2πε0) q ln(1/r) + D E r2(r) = (1/2πε0) q/r region 2 . (L.1.1)' Continuity of φ at r=a and b: (1/2πε 0) q ln(1/b) = (1/2 πα ) q ln(1/b) + C region 0/1 boundary, r = b (1/2πε0) q ln(1/a) + D = (1/2 πα ) q ln(1/a) + C region 2/1 boundary, r = a . (L.1.3)' Rule for E field normal components at a boundary: ε 0Er0(b) = ε1Er1(b) region 0/1 boundary, r = b ε0Er2(a) = ε1Er1(a) region 2/1 boundary, r = a Appendix L : Point and Line Charges in Dielectrics 399 or ε0 q/ [2πε0b] = ε1 (1/2πα ) q/b => 1 = ε1/α ε0 q/ [2πε0a] = ε1 (1/2πα ) q/a => 1 = ε1/α . (L.1.4)' Restated continuity of φ with α = ε 1: (1/2πε 0) q ln(1/b) = (1/2 πε1 ) q ln(1/b) + C region 0/1 boundary, r = b (1/2πε0) q ln(1/a) + D = (1/2 πε1 ) q ln(1/a) + C region 2/1 boundary, r = a . (L.1.5)' Second equation minus first above: D + (1/2 πε 0) q(ln(1/a)- ln(1/b)) = (1/2πε 1)q (ln(1/a)- ln(1/b)) => D + (1/2 πε0) q ln(b/a) = (1/2 πε1) q ln(b/a) . Solution for the three constants: α = ε C = q ln(1/b)(1/2 π) (1/ε 0-1/ε1) D = q ln(a/b)(1/2 π) (1/ε0-1/ε1) . ( L . 1 . 6 ) ' The potentials in the three regions are then φ 0(r) = (1/2πε0) q ln(1/r) φ1(r) = (1/2πε1) q ln(1/r) + (q/2 π) ln(1/b) (1/ε 0-1/ε1) φ2(r) = (1/2πε0) q ln(1/r) + (q/2 π) ln(a/b) (1/ε 0-1/ε1) ( L . 1 . 7 ) ' while the fields are E r0 = (1/2πε0) q/r Er1 = (1/2πε1) q/r Er2 = (1/2πε0) q/r . (L.1.8)' The following Maple plots show the continuity of φ and the jumps in E r at the boundaries Appendix L : Point and Line Charges in Dielectrics 400 Fig L.2' φ(r) [red] and E r(r) [blue] for r in (0.5, 5) What about the (now linear) bound charge densities at r = a and r = b? P = (ε-ε 0)Err^ Pr1(r) = (ε1-ε0)Er1(r) θ(r>a)θ(r<b) // points radially outward since ε 1 > ε0 (L.1.9)' From (1.1.11) the polarization charge density is then ρ pol = - div P (1.1.11) so in cylindrical coordinates, ρ pol(r) = - [r-1∂r(rPr) + r-1∂θPθ + ∂zPz] = - r -1∂r(rPr) = - r -1∂r(r[(ε1-ε0)Er1(r) θ(r>a)θ(r<b)]) = - r-1∂r(r[(ε1-ε0) (1/2πε1) q/r θ(r>a)θ(r<b)]) // from (L.1.8)' = - q( ε1-ε0) (1/2πε1) r-1∂r[θ(r-a)θ (b-r)] = - q( ε 1-ε0) (1/2πε1)r-1 [ δ(r-a) -δ(r-b) ] // from above (L.1.10) = - q( ε 1-ε0) (1/2πε1) [ δ(r-a)/a - δ (r-b)/b ] . (L.1.10)' We may then read off the bound linear charge densities at r = a and b Appendix L : Point and Line Charges in Dielectrics 401 σinner = - (ε1-ε0) (1/2πε1)(q/a) // Coulombs/m σouter = (ε1-ε0) (1/2πε1) ( q / b ) ( L . 1 . 1 1 ) ' so Qinner = ∫{ ds σinner = - (ε1-ε0) (1/2πε1)(q/a) * 2π a = - (1-ε0/ε1) q Qouter = ∫{ ds σouter = (ε1-ε0) (1/2πε1)(q/bb) * 2πb = (1-ε0/ε1) q (L.1.12)' Q = (1- ε0/ε1) q // exactly the same equation as in the 3D case (L.1.13)' Here Q is the total charge/m on the outer surface of the cylindrical shell at r = b, and -Q is the same thing at r = a. Recall that q is the charge/m of the central linear line charge. L.4 Limits of the Previous Problem (a) Line charge in an infinite cylindrical hole in a dielectric Fig L.3' The potentials and fields shown in (L.1.7)' and (L.1.8)' are then, taking b→R (some large value) φ 1(r) = (1/2πε1) q ln(1/r) + (q/2 π) ln(1/R) (1/ε 0-1/ε1) φ2(r) = (1/4πε0) q/r - (q/4 πa)(1/ε0-1/ε1) (L.2.1)' while the fields are E r1 = (1/2πε1) q/r Er2 = (1/2πε0) q/r . (L.2.2)' The induced bound charge density σ at r = a, and the total charge there, are still given by σ = - (ε 1-ε0) (1/2πε1)(q/a) -Q = (1- ε0/ε1) q . ( L . 2 . 3 ) ' Appendix L : Point and Line Charges in Dielectrics 402 (b) Line charge embedded in an infinite dielectric cylinder Fig L.5' The potentials and fields shown in (L.1.7)' and (L.1.8)' are then, taking a → 0 φ0(r) = (1/2πε0) q ln(1/r) φ1(r) = (1/2πε1) q ln(1/r) + (q/2 π) ln(1/b) (1/ε 0-1/ε1) ( L . 2 . 4 ) ' while the fields are E r0 = (1/2πε0) q/r Er1 = (1/2πε1) q/r = (1/2πε0) [q(ε0/ε1)] / r (L.2.5)' where the last expression shows the "shielded charge interpretation". Outside the cylinder, the E field is E r1 = (1/2πε0) q/r, just as if the cylinder were not there. (c) Line charge embedded in an infinite dielectric medium We now take b →R (a large value) in Fig L.5' to remove outer region 0, with this result: Fig L.6' Appendix L : Point and Line Charges in Dielectrics 403 There is only one region left and from (L.2.4)' and (L.2.5') we get φ 1(r) = (1/2πε1) q ln(1/r) + (q/2 π) ln(1/R) (1/ε 0-1/ε1) ( L . 2 . 6 ) ' while the field is Er1 = (1/2πε1) q/r = (1/2πε0) [q(ε0/ε1)] / r (L.2.7)' where the last expression shows the "shielded charge in terpretation". As usual, we can ignore the infinite constant in the potential φ1(r). Appendix M : Why A t is small 404 Appendix M: Why the transverse vector potential A t is small for a transmission line Claim: In the King gauge, the transverse vector potential At may be neglected for frequencies in the range 10f c to 1000 GHz, where f c is the soft loss cutoff frequenc y described below. (M.1) The vector potential is written below as At = ∫conductors dxdy Jt(x,y) * (stuff). We shall make the following three claims: (1) | Jt | < 10-3 |Jz| for f = 10f c Hz to 1000 GHz (which is to say: "transverse currents are small i n s i d e t h e c o n d u c t o r s " ) (M.13) (2) in the At integral there is a cancellation effect not present in the A z integral which in effect reduces At by a factor of 10 (ballpark) relative to A z . (M.14) (3) the net ballpark result is that | At| < 10-4 |Az| for f = 10f c Hz to 1000 GHz which supports the opening claim (M.1) above. (M.22) According to (1.5.9) one can express the King gauge vector potential at all points in space in terms of the currents in the transmission line conductors in this manner, A (x,ω) = 1 4π Σi∫μi Ji(x',ω) e-jβdR R dV' R = | x - x' | (1.5.9) where Σi is a sum over all the conductors, and βd is the wavenumber in the dielectric. Therefore, the transverse part A t may be written as an integral of the transverse conductor currents Jt,i : At(x) = 1 4π Σi μi∫Jt,i(x',y',z') e-jβdR R dx' dy' dz' . (M.2) Transverse refers to the x and y directions, where the infinite conductors are aligned in the z direction. The main current in a transmission line conductor is the longitudinal one J z . When the above equation is processed in the manner of Chapter 4, and one assumes the transmission line limit, the result is At(x) = - 1 4π Σi μi∫Jt,i(x',y') ln(s2) dx' dy' . s2 = (x-x')2 + (y-y'2) (M.3) In this transmission line limit, βd is small (long wavelength) and e-jβdR ≈ 1. We assume the standard wave functional form such that Jt,i(x,y,z) = e-jβdz Jt,i(x,y) and then again set e-jβdz ≈ 1 for the contributing portion of the dz' integrati on and that integration produces -ln(s2) as in Ch 4 or (J.10). Meanwhile, Appendix D computes the E fields inside a round conductor for each partial wave m, and here we multiply them each by σ to get the current density components, Appendix M : Why A t is small 405 Current Densities in a Round Wire: R dc = 1 σπa2 β'2 = β2 - βd2 (D.2.33) Jz(r,m) = σ(1/4) ηm I Rdc (aβ') fm f m = [ Jm(x) Jm+1(xa) - Jm(x) Jm-1(xa) ] x = β 'r Jr(r,m) = σ(j/4) ηm I Rdc (aβd) gm g m = [ Jm+1(x) Jm+1(xa) + Jm-1(x) Jm-1(xa) ] x a = β'a Jθ(r,m) = σ(1/4) ηm I Rdc (aβd) hm h m = [ Jm+1(x) Jm+1(xa) - Jm-1(x) Jm-1(xa) ] The currents are expressed in terms of a cylindrical coordinate system whose z axis runs down the center of the round conductor. Coefficient η m is the "surface charge moment" of the mth partial wave, and the θ- space currents are given by (D.1.3a), J(r,θ) = ∑ m = -∞∞ J(r,m) ejmθ . // partial wave expansion (M.4) The moments ηm may be obtained by solving the transmission line "capacitor problem" as outlined in Section 6.5 (a). One finds potential φ, then E, then surface charge n( θ), and finally ηm. The current components are, J = J zz^ + Jrr^ + Jθ θ^ = Jzz^ + Jt Jt = Jrr^ + Jθ θ^ . (M.5) Rather than study these round wire internal solutions in detail, we make two observations: Observation (1): The transverse currents J r and Jθ are very small compared to J z. (M.6) Looking at (D.2.33) above one sees that the transverse currents J r and Jθ are in general smaller than the longitudinal current J z by factor | βd/β'|. In making this claim we rega rd the combinations of Bessel functions shown in (D.2.33) as being of the same gene ral scale, which can be confirmed by doing plots of the various complex function magnitudes. Consider these relations : ( first line is from (1.5.1a), (3.3.2) with ε ' ≈ ε, and (1.1.29) that v d = 1/με ) βd = βd0 [1 - j (1/2)tan L] ≈ βd0 = (ω/vd) = 2π/λd ( M . 7 ) β = ej3π/4 (2 /δ) = [(j-1)/ 2 ] ( 2 /δ) = (j-1)(1/ δ) δ ≡ 2/(ωμσ) (2.2.21) (M.8) β'2 = β2 - βd2 . (D.2.2) (M.9) Appendix M : Why A t is small 406 Using these facts there are several ways to write the ratio | βd/β |, one of which is this: => | βd/β | = 2πδ /(2 λd) = 2 π (δ/λd) ( M . 1 0 ) which at least suggests that | βd/β | is small since one normally thinks of skin depth δ as being much less than the wavelength of a wave on the transmission line. As shown in Appendix D.11, when losses are included one really has β d = (ω-jωc)/vd where ωc = 2πfc is a soft cutoff frequency below which line losses become into lerable. For Belden 8281 coaxial cable it is shown that f c ≈ 7 KHz, while for a typical power transmission line f c ≈ 5 Hz. Since one would never operate a transmission line with ω < ωc, that frequency region is of little interest to us. A healthy lower limit might be ω = 10ωc. A more useful expression of the ratio| β d/β | is the following, where we use fact (1.1.2) that εμ = 1/vd2 and we include the loss effect just mentioned, | βd β |2 = |ω-jωc vd |2 δ2/2 = |ω-jωc vd |2 1 ωμσ = | ω-jωc|2 με 1 ωμσ = | ω-jωc|2 ε ωσ = | 1-jω c/ω|2 εω σ => | βd/β | = | 1-j ωc/ω| εω/σ . ( M . 1 1 ) At some low frequency (perhaps ω = 10ω c) this ratio is fairly small (being ~ εω/σ ), but it then increases with ω as ω1/2. If we are willing to restrict our transmission line interests to 10f c < f < 1000 GHz, and if the conductors are copper, we find that the ratio | βd/β | will always be less than the following: εω/σ ≈ ε0ω/σ = ε02πf/σ = [8.85 x 10-12 * 2π * 1012 / 5.81 x 107 ]1/2 = 10-3 (M.12) Thus, all the way from f = 10f c to f = 1000 GHz, we have | βd/β| < 10-3. At 10 GHz the ratio is 10-4. Since | β| << |βd|, it follows that β'2 = β2 - βd2 ≈ β2 so then β ' = β for all practical purposes and then we have shown that | βd/β'| << 10-3. Thus from (D.2.33) quoted above, | J r | < 10-3 | Jz | | Jθ | < 10-3 | Jz | f = 10f c to 1000 GHz (M.13) The conclusion then is that the tran sverse currents are less than 1/1000th of the size of the longitudinal currents for the round copper conductor at all frequ encies of interest below 1000 GHz, and we can reasonably assume that a similar conclusion applies to a conductor of any cross sectional shape. This then concludes our "proof" of the claim that "transverse currents are very small" inside the conductors of a transmission line." Observation (2): In the Helmholtz integration (M.3) ther e is a large amount of cancellation. (M.14) Appendix M : Why A t is small 407 Let us consider the nature of this integration in the illustrative case of a two round conductors, Fig M.1 Consider the contribution to the transverse vector potential component A x from the right conductor C 2, A x(x) = - μd 4π ∫Jx(x'2,y'2) ln(s22) dx'2 dy'2 s 22 = (x-x'2)2 + (y-y'22) (M.15) or A x(x) = - μd 4π ∫ [Jr(r'2,θ'2) r^'2 • x^ + Jθ(r'2,θ'2) θ^'2 • x^ ] ln(s22) [a2dθ'2] dr'2 (M.16) or A x(x) = - μd 4π ∫ [Jr(r'2,θ'2) cosθ'2 - Jθ(r'2,θ'2) sinθ'2] ln(s22) [a2dθ'2] dr'2 . (M.17) The transverse currents in conductor C 2 have this partial wave expansion from (M.4), Jr(r'2,θ'2) =∑ m = -∞∞ Jr(r'2,m) ejmθ'2 ( M . 1 8 ) and similarly for J θ . Thus we get Ax(x,y) = - μd 4π a2 Σm ∫0a2 dr'2 Jr(r'2,m) ∫02π dθ '2 ejmθ'2 cosθ'2 ln(s22) + μd 4π a2 Σm ∫0a2 dr'2 Jθ(r'2,m) ∫02π dθ '2 ejmθ'2 sinθ'2 ln(s22) (M.19) where, from (D.2.33) quoted above, Appendix M : Why A t is small 408 Jr(r,m) = σ(j/4) ηm I Rdc (aβd) [ Jm+1(x) Jm+1(xa) + Jm-1(x) Jm-1(xa) ] x = β 'r Jθ(r,m) = σ(1/4) ηm I Rdc (aβd) [ Jm+1(x) Jm+1(xa) - Jm-1(x) Jm-1(xa) ] x a = β'a . It is in theory possible to first do the d θ'2 integration in (M.19) and then do the dr' 2 integration and get an analytic result for A x(x,y). We have dealt with similar angle integrations elsewhere in this document. Rather then attempt this task, we instead consider the portion of the 2D integration represented by the red ring in Fig M.1. On this ring, r' 2 is constant, and our interest is the θ'2 integration. For any value of m (except for ±1) the trigonometric functions like ejmθ'2 cosθ'2 integrate to 0, for example, F i g M . 2 For these values of m, were it not for the fact that s 22 varies around the red circle, A x(x,y) would be identically 0. Although s 22 does vary on the red circle, ln(s2) varies very little, and we expect to still have this strong cancellation in the θ'2 integral so A x(x,y) is then small. It is true that if x and x'2 were to approach the conductor boundary from opposite sides, then ln(s2) would vary a lot more and the cancellation would be less, but we ignore this detail in our qualitative argument. For m = ± 1 this smallness argument fails since for example cos2(θ'2) does not average to 0 around the red ring. Ignoring the θ'2 variation in s 22 we get in this case (setting e±jθ'2 ~ cosθ'2) Ax(x,y) ≈ - μd 4π a2 Σm ∫0a2 dr'2 Jr(r'2,m) ln(s22) ∫02π dθ '2 cos2θ'2 ≈ - μd 4π a2 Σm ∫0a2 dr'2 Jr(r'2,±1) ln(s 22) π ( M . 2 0 ) Now we make a different argument which concerns th e behavior of the complex Bessel functions as a function of r' 2. As studied in Chapter 2, these functions have a dramatically oscillating phase even in the soft skin depth limit, and we expect then to get can cellation due to this phase as we integrate on the radial segment shown blue in Fig M.1, and again ln(s 22) varies slowly on this ray due to the nature of ln. The arguments made above for A x(x,y) also apply to A y(x,y), and it seems reasonable to assume that the arguments are generally valid for an arbitrary conductor cross section. Admittedly our analysis here is imprecise and qualitative, but we think it is convincing that there is in fact much cancellation when the transverse cu rrents are integrated over the conductors. Appendix M : Why A t is small 409 This stands in stark contrast to the longitudinal situation where A z, being the Helmholtz integral of J z, involves a generally non-cancelling integration (per conductor) over a generally large current component. We now wish to compare the following two integrals, where we pick component A r to represent a transverse component of At, Ar(x) = - 1 4π Σi μi∫Jr,i(x',y') ln(s2) dx' dy' . s2 = (x-x')2 + (y-y'2) (M.3) Az(x) = - 1 4π Σi μi∫Jz,i(x',y') ln(s2) dx' dy' . s2 = (x-x')2 + (y-y'2) (M.21) We have shown in (M.13) that | J r | < 10-3 | Jz |. Without any mathematical rigor, and allowing a factor of 10 "gain" from the cancellation effect of Observation (2), we make the following ballpark estimate, |At| < 10-4 |Az| f = 10f c to 1000 GHz. (M.22) It is assumed that as ω increases, the transmissi on line geometry is appropriately shrunk so the transmission line limit remains operative. Appendix N : Magnetic Ohm's Law and the Radial Hall Effect 410 Appendix N: Drude, Magnetic Ohm's Law, Regular Hall Effect, Radial Hall Effect The first sect ions of this Appendix follow the general outline of notes prepared by Pengra et. al. for a Laboratory Class at the University of Washington. N.1 The Drude Model of Conduction The current due to carriers of charge q and density n with drift velocity v is easily shown to be J = nq v . dim RHS = m-3 * Coul * m/sec = amp/m2 (N.1.1) In the classical 1900 Drude/Lorentz model (which of course predates quantum mechanics), the charges are assumed to be electrons with charge q = - |e| and mass m = m e. At this time there was no band-gap theory, no holes, no effective mass, none of that good stuff. The density n is one electron per atom for a metal like copper. Here are some basic numbers : n = 8.5 x 1028 electrons/m3 // for copper |e| = 1.6 x 10 -19 C o u l . ( N . 1 . 2 ) If a relatively large current of 1000 Amps flows through a wire of 1 cm 2 cross sectional area, one has J = 1000 amps/ 10 -4m2 = 107 amp/m2 . The drift velocity is then v = J/(nq) = 100 1.6*8.5 x 105+19-28 = 7.4 x 10-4 m/sec = 0.74 mm/sec ≈ 1 mm/sec In this same classical vein, if the electron has thermal energy (1/2) mv th2 = (3/2) kT, on e can solve for the thermal electron velocity at room temperature, vth ~ 100,000 m/sec . Although this number is wrong from a quantum view, the f act that it is very much larger than the drift velocity is correct. In the Drude theory, these fast-moving electrons are colliding with copper ions at a high rate, and every collision results in a complete re direction of the electron. In copper the effective mean collision time is on the order of τ = 10 -14 sec. It is only between these closely spaced collisions that the electrons have time to drift a little in the presence of an electric field. Since F = qE = d p/dt, one concludes that Δp = qEΔt or just p = qEτ where p is the amount of drift momentum an electron picks up between collisions. Since on average an electron on each collision dumps this momentum into the lattice, the lattice can be regarded as a frictional or drag forc e acting against the electron's flow, and that force is - Δp/Δt = - p /τ . So, Ff = - p/τ = - (m/τ) v . ( N . 1 . 3 ) Appendix N : Magnetic Ohm's Law and the Radial Hall Effect 411 This frictional force is proportional to velocity, as is typical for low-veloc ity fluid drag, and is naturally in a direction opposite the velocity. When combined with the Lorentz force, F = qE + q vxB ( N . 1 . 4 ) and F = ma, one obtains a fairly reasonable equation d escribing the motion of a conduction electron, mdv dt = qE + q vxB - (m/τ ) v . ( N . 1 . 5 ) If B = 0 and the conduction is in steady-state, this says 0 = q E - (m/τ ) v or v = (qτ/m)E . ( N . 1 . 6 ) The constant appearing he re is called the carrier mobility μ, so then v = μE μ = (qτ/m) . // units of μ are tesla -1 (N.1.7) Officially mobility is (|q| τ/m) > 0, but we shall use the signed mobility shown above. Warning : μ is the same symbol used for magnetic permeability. If one now installs the drift veloci ty (N.1.6) into (N.1), one gets J = nq v = (nq 2τ/m)E = σE . σ = conductivity (N.1.8) The coefficient appearing in (N.1.8) is known as the conductivity of the medium, as we well know by now, so the classical Drude theory is predicting that σ = (nq 2τ/m) . // σ = n q μ ( N . 1 . 9 ) If one measures σ for copper, one can deduce the value of τ for the Drude model of conduction: τ = mσ/(nq2) ( N . 1 . 1 0 ) We know that m = 9.109 x 10-31 kg // electron mass σ = 5.81 x 107 mho/m // conductivity of copper (N.1.11) so that, along with the numbers stated earlier in (N.1.2), τ = mσ/(nq 2) = 2.43 x 10-14 ~ 10-14 ( N . 1 . 1 2 ) Appendix N : Magnetic Ohm's Law and the Radial Hall Effect 412 as claimed earlier. If the electrons are moving with time dependence ejωt, the left side of the equation of motion (N.1.5) becomes j ωm v. We then get jωm v = qE- (m/τ ) v (m/τ)(1+jωτ) v = qE v = 1 1+jωτ (qτ/m)E μ ac = 1 1+jωτ μ ( N . 1 . 1 3 ) J = nq v = 1 1+jωτ (nq2τ/m)E = σacE σac = 1 1+jωτ σ . (N.1.14) In our analysis of transmission lines, ωτ << 1, so we may neglect this AC adjustment of the mobility and conductivity. Roughly ωτ ≈ 1 when ω = 2πf = 1/τ = 10 14 => f ≈ 16,000 GHz (N.1.15) so for f < 160 GHz there will be < 1% change in μ or σ in the Drude Model. N.2 A Theory of the Hall Effect All theories and m odels are deficient in some way but might still deliver a reasonable result. The Drude model above is generally "reasonable" in this regard, though it fails to match reality in various ways. Here we present an instant theory of the Hall Effect which correctly predicts the main result to within about 30%, but has an annoying theoretical defect noted at the end of the section. Using the traditional directions x, y, and z, here is the classical Hall Effect picture, where we put the origin at the center of the sample, Fig N.1 The idea is that current flows through a sample in the presence of a uniform transverse magnetic field which in this case is B = Bzz^ . Semiconductors have much lower carrier densities than copper, and one can imagine for a semiconductor sample that the bl ock above is placed betw een two highly conductive gold plates (gray on right) to cause the applied current to be spread out evenly in the sample. This is one Appendix N : Magnetic Ohm's Law and the Radial Hall Effect 413 of several technical details we shall ignore, and we just assume the current is spread out evenly. Typically the thickness T is made very small because this boosts the Hall voltage as we shall see below. For the sake of our discussion, we assume we are in an anti-matter universe where the carriers are positive electrons (positrons) and the lattice consists of negative ions. We just want to deal first with positively charged carriers since then v x and Jx and I are all positive. So assume q > 0. The Lorentz force acting on a carrier of charge q, along with the friction term, was shown in (N.1.5), F = q E +q vxB - (m/τ ) v . (N.1.5) One's right hand indicates that v xB is downward in the -y direction, so there is a force deflecting the positive carriers downward, and so for a while there is some downward v y drift. This naturally piles up a positive surface charge on the lower face of the sample. Since the density of positrons and anti-copper ions must be the same to maintain neutrality in th e copper sample, these positrons in effect come from the upper face which then has a negative surface charge. Thus we have in effect a parallel plate capacitor, positive on the bottom, with spacing W, and a transverse field E y > 0 appears due to these surface charges. This E y field then stops further downward deflec tions, and the positrons then have only v = vx^ . The steady-state solution value of E y is determined by Fy = qEy + q (v xB)y - (m/τ) vy = 0 But v y = 0 so this says Ey = - (vxB)y = - ( vxB)•y^ = - { [v xx^] x [Bzz^ ]} • y^ = vxBzy^ • y^ = vxBz or Ey = vxBz . // the Hall field (N.2.1) From (N.1.1) we have J x = nqvx and also J x = I/(WT) so vx = I nqWT and then E y = IBz nqWT . (N.2.2) One usually then defines R H ≡ 1 nq // the Hall coefficient (N.2.3) so (N.2.2) becomes vx = RH I WT and E y = RH IBz WT // the Hall field (N.2.4) This Ey field produces a potential (a voltage) between the top and bottom faces, and since E = - ∇V, Appendix N : Magnetic Ohm's Law and the Radial Hall Effect 414 VH = Vtop - Vbot = V(W) -V(0) = ∫0 W dV dy dy = – ∫0 W Ey dy = - E y W = - R H IBz WT * W so VH = – RH * IBz T // the Hall voltage (N.2.5) We now return the reader to the regular universe. Ho w is the above discussion altered? Since the carriers are now electrons with q = -|e| < 0, the v x arrow in Fig N.1 points to the left and we have v x < 0. From (N.2.2) the Hall field E y changes sign, becoming negative (though it is still E y = vxBz ), and the Hall voltage changes polarity and is now positive ( th e Hall coefficient is negative). To summarize: RH = 1 nq = - 1 n|e| < 0 Ey = RH IBz WT = – 1 n|e| IBz WT < 0 VH = – RH IBz T = + 1 n|e| IBz T > 0 (N.2.6) Notice that the electrons are still deflected down (as were the positrons) because both q and v x B change sign in the Lorentz force. This now creates negative ch arge on the bottom face a nd positive on the top and so now E y < 0, consistent with E y = vxBz with vx < 0. Fact : The sign of the Hall voltage V H indicates the sign of R H and thus the sign of the charge carriers! If for some metal the carriers are holes ( in the quantum theory of metals), R H will be positive. Pre-quantum researchers were indeed surprised when they found different signs of R H for different metals. Using the numbers in (N.1.2), the Drude theory for copper predicts that RH ≈ -.73 x 10-3 / / D r u d e t h e o r y ( N . 2 . 7 ) Here is the simple Maple calculation using numbers from (N.1.2) above, This is not too far from the measured a nd quantum-correct value of -0.55 x 10-10 (though the literature seems a bit unsure of this number). This is an impressive success of the classical Drude theory. Appendix N : Magnetic Ohm's Law and the Radial Hall Effect 415 As claimed earlier, and as seen in (N.2.5), making thickness T very small makes V H larger so it can be measured with a voltmeter one can afford to place in a student lab. Typical numbers for a student lab experiment might be T = 18 microns = 18 x 10 -6 m // a thin film of copper W = 1 cm = 1 x10-2m Bz = 5000 gauss = 0.5 T I = 1 0 a m p s ( N . 2 . 8 ) so that, according to the Drude theory, so we end up for this experiment with E y = - 2 mV/m Hall field VH = 20 μV Hall voltage vx = - 0.4 mm/sec drift velocity (N.2.9) Notice that R H = - 1 n|e| . Since the carrier densities in a semiconductor are much smaller than in a metal, n is smaller and R H is much larger, and practical Hall devices are more feasible. But our theory has to first be generalized to two types of carriers (electrons and holes), and this is done in Section N.6 below. A Hall effect sensor exists in almost every fan in every personal computer in the world. Since the fan has some rotating permanent magnets, the Hall sensor can detect the rotational position and speed of the blades and most importantly detects when the fan has stopped rotating altogether (pulses stop). In general, Hall sensors are used to measure magnetic fields, and can be used as simple magnetic switches. If we assume that Ohm's Law J = σ E is operative in our Hall sample, we run into a small deficiency in the theory. Since there is an internal field E y in the sample, there should be a corresponding and uniform current density J y = σ Ey in the sample. Unfortunately, at th e top face (for example) this current has no place to go, so something is wrong. This problem will be dealt with in Section N.5 below. Appendix N : Magnetic Ohm's Law and the Radial Hall Effect 416 N.3 The Cyclotron Frequency When a charged particle travels through an em pty region of space having a uniform B field and no E field, the equation of motion (N.1.5) becomes (dot means time derivative), m v• = q vxB . (N.1.5) Assume that B = B z^ so then q vxB = [v xx^ + vyy^ + vzz^] x [qB z^] = - vxqBy^ + vyqBx^ so mv•x = vyqB => v•x = ωcvy => v••x = -ωc2vx mv•y = -vxqB v•y = - ωcvx v••y = -ωc2vy mv•z = 0 v•z = 0 where ωc ≡ (qB/m) . (N.3.1) Looking at the 2nd order ODE for v x we may write the general solution for v x in terms of two constants R and φ in this way vx = -Rωcsin(ωct + φ ) => v y = (1/ωc) v•x = - Rωccos(ωct + φ ) so that v = vx2+vy2 = Rωc = ( R q B / m ) ( N . 3 . 2 ) vx = - Rωcsin(ωct + φ ) => x = R cos(ω ct + φ ) + x1 vy = - Rωccos(ωct + φ ) => y = -R sin( ωct + φ ) + y1 vz = vz => z = v zt + z0 (N.3.3) so (x-x 1) = R cos( ωct + φ ) (y-y 1) = - R sin( ωct + φ ) => (x-x 1)2 + (y-y1)2 = R2 (z-z0) = vzt . ( N . 3 . 4 ) In the x,y dimension the particle goes ar ound in a circle of radius R at rate ω c (clockwise if ωc > 0), while in the z direction of B it moves at some constant velocity, resulting in a circular or slinky spiral trajectory. The angular frequency ω c is known as the cyclotron frequency, named after a charged-particle accelerator invented in 1932 by Lawrence known as a cyclotron, see wiki and left drawing below. In this machine particles traverse an outward going spiral (d ifferent from the one just mentioned) because they are accelerated by an AC electric field driving two hollow D-shaped conductors of a capacitor enclosing the particle beam. As v increases, R must increase as s hown above in (N.3.2). The capacitor is driven at the cyclotron frequency ω c so the accelerating E field is in sync with the circular particle motion. Since ωc ≡ (qB/m), this frequency has to be reduced if the charged particle bunch being accelerated reaches relativistic speeds and m increases. Appendix N : Magnetic Ohm's Law and the Radial Hall Effect 417 The circular motion of charged particles in a uni form B field is also used to identify particles produced in high energy collisions inside particle acceler ator detectors. Since v = RqB/m, if the particle m and q is known, the speed v and hence energy can be determined from R, and the sign of q can be found from the CW or CCW nature of the particle path. Al ternatively, if the energy and speed are known from "calorimetry" and a charge q is assumed, the mass m of the particle can be found from R. http://hyperphysics.phy -astr.gsu.edu/hbase/magnetic/cyclot.html CERN The Cyclotron Particle Tracks (B field out of paper) Fig N.2 N.4 Steady-state Electron Motion with E and B fields: Magnetic Ohm' s Law We start again with the motion equation for an electron in copper, mdv dt = qE + q vxB - (m/τ ) v . (N.1.5) We now seek a steady-state solution, so the equation becomes (m/τ) v = qE + qvxB . ( N . 4 . 1 ) Making use of the signed mobility μ = qτ/m shown in (N.1.7), we can write (N.4.1) as v = μE + μvxB or v - μvxB = μE . (N.4.2) The plan is to solve this equation for v and then to obtain the current density using J = nq v from (N.1.1). Recall that Ohm's Law says J = σ E, but with B present, Ohm's Law will be different. For simplicity, we again assume B = B z^. Then μ vxB = [v xx^ + vyy^ + vzz^] x [μ B z^] = - vxμBy^ + vyμBx^ Appendix N : Magnetic Ohm's Law and the Radial Hall Effect 418 so that (N.4.2) becomes [vxx^ + vyy^ + vzz^] - [- vxμBy^ + vyμBx^ ] = [μExx^ + μEyy^ + μEzz^ ] which may be decomposed into the following three equations, v x - μBvy = μEx vy + μBvx = μEy vz = μEz . ( N . 4 . 3 ) The first two equations may be expressed in matrix form ⎝⎛ ⎠⎞ 1 -μB μB 1 ⎝⎛ ⎠⎞ vx vy = μ ⎝⎛ ⎠⎞ Ex Ey ( N . 4 . 4 ) and then ⎝⎛ ⎠⎞ vx vy = μ ⎝⎛ ⎠⎞ 1 -μB μB 1 -1 ⎝⎛ ⎠⎞ Ex Ey . (N.4.5) Maple tells us so then ⎝⎛ ⎠⎞ vx vy = μ 1+(μB)2 ⎝⎛ ⎠⎞ 1 μ B -μB 1 ⎝⎛ ⎠⎞ Ex Ey = μ 1+(μB)2 ⎝⎛ ⎠⎞ Ex + μBEy Ey - μBEx . (N.4.6) But using (N.3.1) that ωc ≡ (qB/m) one finds that μB = (qτ/m)B = (qB/m) τ = ω cτ ( N . 4 . 7 ) so the solutions above, combined with the known third solution v z = μEz, become Appendix N : Magnetic Ohm's Law and the Radial Hall Effect 419 vx = μ( Ex + ωcτ Ey) 1 1+(ωcτ)2 vy = μ (Ey - ωcτ Ex) 1 1+(ωcτ)2 vz = μEz . ( N . 4 . 8 ) We have found our solution for v ! To find J use (N.1.1) that J = nq v and the fact that nqμ = nq(qτ/m) = (nq2τ/m) = σ // from (N.1.7) and (N.1.9) (N.4.9) to find that (in agreement with (10) of Pengra), Jx = σ (Ex + ωcτ Ey) 1 1+(ωcτ)2 ωc ≡ (qB/m) Jy = σ (Ey - ωcτ Ex) 1 1+(ωcτ)2 B = B z^ Jz = σEz . σ = (nq2τ/m) (N.4.10) The is the "Magnetic Ohm's Law" which, in the presence of B = B z^ , replaces the usual Ohm's Law, Jx = σEx Jy = σEy Jz = σEz (1.1.7) Fortunately, as will be shown below, the magnetic field strength in a transmission line is small enough so that ωcτ << 1, which means that the normal Ohm's Law is justified despite the presence of B fields. N.5 Theory of the Hall Effect Revisited We replicate the Hall geometry from above, where recall that B = Bzz^ : Fig N.1 Looking at the drawing, and recalling the small "defect" in the theory of Section N.3, and staring at the Magnetic Ohm's Law (N.4.10), we insist that J y = 0 at least at the top and bottom faces, since as noted earlier, this current "has nowhere to go " in the steady state. Since things are generally uniform in this slab of material, we make the ansatz that J y ≡ 0 everywhere in the sample. The second equation of (N.4.10) then says Appendix N : Magnetic Ohm's Law and the Radial Hall Effect 420 Ey - ωcτ Ex = 0 ( N . 5 . 1 ) and when this is inserted into the first equation we find, along with the other two equations of (N.4.10), Jx = σ (Ex + ωcτ [ωcτ Ex]) 1 1+(ωcτ)2 = σ Ex Jy = 0 Jz = σEz . ( N . 5 . 2 ) There is no reason to have E z ≠ 0 since the electrons are only deflected up and down. Moreover, we would like to have J z = 0 on the front and back face, so E z ≡ 0 is the obvious choice. Then from (N.5.1) we must have Ey = ωcτ Ex = ωcτ [Jx/σ] = ωcτ Jx σ * I JxTW * qBz/m ωc * σ nq2τ/m = τ 1 * I TW * qBz/m 1 * 1 nq2τ/m = 1 nq * I Bz TW . ( N . 5 . 3 ) This is the Hall field ! The Hall voltage is then V H = – EyW = 1 nq * I Bz T = – RH I Bz T R H = 1 nq . (N.5.4) This is the same as the Hall voltage obtained in our previous derivation, as shown in (N.2.4) and (N.2.5). But now we end up with J y = 0 so there is no vertical current ha ving nowhere to go, nor is there front- back current, and we also have J x doing the regular Ohm's Law as shown in (N.5.2) Jx = σEx Jy = 0 Jz = 0 . ( N . 5 . 5 ) This seems a more complete solution to the Hall problem than that of Section N.2. N.6 Theory of the Hall Effect with Multiple Carrier Types Let index i label the ty pes of carriers. The devel opments of Sections N.1 through Section N.4 carry through as is, but everything now has an i index. For example, we now have Appendix N : Magnetic Ohm's Law and the Radial Hall Effect 421 J = Σiniqivi (N.1.1) (N.6.1) vi = (qiτi/mi)E = μi E μi = (qiτi/mi) = signed mobility (N.1.7) (N.6.2) σ i = niqi μi = niq (qiτi/mi) = (niqi2τi/mi) (N.1.9) (N.6.3) v i - μivixB = μiE . (N.4.2) (N.6.4) This leads to solutions for velocities v i , vxi = μi( Ex + ωciτi Ey) 1 1+( ωciτi)2 ωci ≡ (qiB/mi) vyi = μi (Ey - ωciτi Ex) 1 1+( ωciτi)2 μi = (qiτi/mi) vzi = μi Ez (N.4.8) (N.6.5) and we can define the total conductivity as σ ≡ Σ i σi . ( N . 6 . 6 ) The current densities from (N.6.1) and (N.6.5) are then, using also (N.6.3) that σ i = niqi μi , Jx = Σi σi [ ( Ex + ωciτi Ey) 1 1+( ωciτi)2 ] ωci ≡ (qiB/mi) Jy = Σi σi [ (Ey - ωciτi Ex) 1 1+( ωciτi)2 ] ωciτi = (qiτiB/mi) = Bμi Jz = Σi σi Ez = Ez Σi σi = Ez σ . (N.6.7) At this point it is useful to define objects α and β having the dimensions of conductivity, and a third object which is γ = β/B : α ≡ Σi σi 1 1+( ωciτi)2 β ≡ Σi σi ωciτi 1+( ωciτi)2 = B Σi σi μi 1+( ωciτi)2 = B γ γ ≡ Σi σi μi 1+( ωciτi)2 . (N.6.8) In terms of α and β we rewrite (N.6.7) as Jx = α Ex + βEy Jy = α Ey - βEx Jz = Ez (Σi σi) = Ez σ . ( N . 6 . 9 ) Appendix N : Magnetic Ohm's Law and the Radial Hall Effect 422 Our Hall effect geometry again requires that J y = 0 and that J z = 0 ("nowhere to go") , Fig N.1 so the second equation of (N.6.9) says α Ey = βEx or Ey = (β /α) Ex // = the Hall field (N.6.10) and this is the Hall-effect electric field in the case of multiple carrier types. Inserting this into the first equation of (N.6.9) gives J x = α Ex + βEy = α Ex + β (β/α) Ex = [ α + β2/α ] Ex ≡ σmr Ex (N.6.11) so our triplet of current densities is now J x = [α + β2/α] Ex Jy = 0 Jz = 0 . ( N . 6 . 1 2 ) The conductivity appearing in the J x equation we might define as σmr so that Jx = σmr Ex σmr = α + β2/α = α(B) + B2 γ(B)2/α(B) (N.6.13) where we must remember that α , β and γ all depend on B through each ωciτi = Bμi. In general, we have σmr ≠ σ, so the Hall sample has a conductivity in the main current direction x which depends in a complicated manner on field B. This effect is called magnetoresistance . However, if there is only one carrier type, one finds that σmr = σ (see below) and there is then no magnetoresistance effect, as we already saw in the first equation of (N.5.2). The Hall field can be written Ey = (β /α) Ex = (β/α) [ α + β2/α ]-1 Jx = (β/α) [ α + β2/α ]-1 I / (WT) = β α2+β2 I / (WT) = Bγ α2+Bγ2 I / (WT) = γ α2+Bγ2 B I / (WT) (N.6.14) and then the Hall voltage is Appendix N : Magnetic Ohm's Law and the Radial Hall Effect 423 VH = - EyW = - γ α2+B2γ2 B I / T ( N . 6 . 1 5 ) and the Hall coefficient is R H = γ α2+β2 = γ(B) [α(B)]2+[β(B)]2 . ( N . 6 . 1 6 ) Unlike the single-carrier case, R H now depends on B in a complicated manner. Just to verify the single carrier case we evaluate: α = σ 1 1+(ωcτ)2 β = σ ωcτ 1+(ωcτ)2 γ = σ μ 1 1+(ωcτ)2 α2 + β2 = σ2 1 1+( ωcτ)2 RH = γ α2+β2 = σμ σ2 = μ/σ = (qτ/m) / (nq2τ/m) = 1/(nq) σmr = α + β2/α = (1/α )( α2 + β2) = σ2/σ = σ . // no magnetoresistance (N.6.17) If we take the magnetic field B small enough so that ω ciτi << 1 for all carrier types, where recall that ωci ≡ (qiB/mi), there is considerable simplification. We find that α ≡ Σi σi 1 1+( ωciτi)2 ≈ Σi σi = σ β = Σi σi ωciτi 1+( ωciτi)2 ≈ Σi σi ωciτi 1 << Σi σi 1 << σ α2 + β2 ≈ α2 ≈ σ2 γ = Σi σi μi 1+( ωciτi)2 ≈ Σi σiμi . // signed mobilities (N.6.18) Then we find that R H ≈ Σi σiμi σ2 = Σi σiμi (Σi σi)2 . ( N . 6 . 1 9 ) For two charge carrier types this gives R H = σ1μ1 + σ2μ2 (σ1+σ2)2 = n1q1 μ1 μ1 + n2q2 μ2 μ2 (n1q1 μ1 + n2q2 μ2)2 . (N.6.20) Now suppose 1 = hole and 2 = electron so then q 2 = -q1. Then we get, where q 1 = |e|, Appendix N : Magnetic Ohm's Law and the Radial Hall Effect 424 RH = 1 |e| → 1 |e| n1μ12 - n2μ22 (n1μ1 + n2μ2)2 // weak B field (N.6.21) where in the last form we revert to the official a ll-positive mobilities. This result is in agreement with Eq. (13) of our Pengra et. al. reference. The Hall coefficient could have either sign! N.7 The Radial Hall Effect in a Round Wire The author ha s had difficulty finding a treatment of this subject, but it must exist somewhere. Consider an "isolated" infinitely long round wire of ra dius a carrying static current I. We use cylindrical coordinates r, θ,z with the symmetry axis along the wire center line. It is often casually claimed that the current density J z in such a wire is uniform throughout the cross section and that there is no charge density on the surface. Here we wish to explore these claims. In cross section, the situation is as follows: Fig N.2 Each electron sees the magnetic field B created by all the other flow ing electrons. At any azimuthal location θ, the electrons are deflected to ward the center line by this B field, causing a free charge distribution inside the wire which results in a radial field component E r. This radial Hall field then offsets the deflection resulting in all electrons flowing exactly in the z direction. This problem differs from the regular Hall effect problem studied in Sections N.2 and N.5 in two major ways: (1) The magnetic field is generated by the flowing current under study, it is not externally applied; (2) the magnetic field is non-uniform and in fact is a function of r. We shall use the method of Section N.5 to determine the E r field and the associated charge distribution. The three unit vectors x^, y^, z^ of that section can be replaced by the cylindrical unit vectors r^, θ^, z^ where the usual cyclic sense of unit vector cross produc ts is then maintained. In a cross section of the round wire, r and θ are then "the usual" polar coordinates, while the z axis comes out of the plane of paper. Since the situation is static, one must have curl E = 0 . But in cylindrical coordinates, Appendix N : Magnetic Ohm's Law and the Radial Hall Effect 425 curl E = r^ [ r-1∂θEz - ∂zEθ] + θ^ [∂zEr - ∂rEz] + z^ [ r-1∂r(rEθ) - r-1∂θEr ] . 1 2 3 4 5 6 We certainly expect to have E θ = 0 at r = a since the round wire surface should be an electrostatic equipotential, and it seems reasonable to have E θ = 0 everywhere inside the wire, so we set E θ= 0 as an ansatz in a search for a Maxwell-Equations solution, and this knocks out terms 2 and 5 Any term with ∂z must also vanish since the wire is static and infinite in length, killing off terms 2 and 3. Since the wire is in isolation, the field pattern must be azimuthally symmetric, so ∂θ terms vanish, killing off 1and 6. Having thus removed terms 1,2,3,5,6, we are left only with term 4 so curl E = θ^ [- ∂rEz] . ( N . 7 . 1 ) Since the static situation requires curl E = 0, we end up with ∂rEz(r,θ,z) = ∂rEz(r) = 0 => E z(r) = constant . (N.7.2) Recalling from Section N.4 that Ohm's Law can be affected by magnetic fields, we now make a second ansatz which is that the regular Ohm's Law a pplies in the z direction. We then obtain Jz(r) = σ Ez(r) = constant J z = uniform (N.7.3) and in this way we arrive at a uniform J z in the wire, but we need to verify that our assumptions made so far are consistent with other requirements. Given then that J z is constant in r and θ, we can compute the magnetic field inside the wire from Ampere's Law in the usual fashion, 2πr H(r) = πr2 πa2 I => H(r) = I 2πa (r/a) => B(r) = μ0I 2πa (r/a) θ^ = B(r) θ^ (N.7.4) Here we make a third ansatz that the other two B field components are 0. Note that μ0 is magnetic permeability, while μ to appear below is the (signed) electron mobility. At this point, we recall the static equation (N.4.2) arising from the Lorentz force and collision friction, v - μvxB = μE (N.4.2) (N.7.5) and we solve for v using the method of Section N.4. First, μ vxB = [v r r^ + vθ θ^ + vz z^] x [μ B θ^] = vr μ B z^ - vz μ B r^ v = v r r^ + vθ θ^ + vz z^ Then (N.7.5) becomes Appendix N : Magnetic Ohm's Law and the Radial Hall Effect 426 [vr r^ + vθ θ^ + vz z^] - [vr μ B z^ - vz μ B r^] = μEr r^ + μEθ θ^ + μEz z^ which may be decomposed into the following three equations ( here in z,r,θ order), vz - μ B vr = μEz vr + μ B vz = μEr vθ = μEθ . ( N . 7 . 6 ) We note that the first two equations of (N.7.6) have the same form as the first two equations in (N.4.3) which were v x - μBvy = μEx vy + μBvx = μEy (N.4.3) Taking then the previous solution with (x,y) → (z,r) we find from (N.4.8) that v z = μ( Ez + ωcτ Er) 1 1+(ωcτ)2 vr = μ (Er - ωcτ Ez) 1 1+(ωcτ)2 vθ = μEθ . ( N . 7 . 7 ) where we have carried down the third equation from above. Recall that the cyclotron frequency ωc enters the picture since μB = ωcτ as shown in (N.4.7). However, now since B = B(r), we have ωc = ωc(r) = qB(r)/m. Nothing in the development of Section N.4 precluded the B field from having spatial dependence because no spatial derivatives (like curl or div) were involved. The next step is to use (N.1.1) that J = nq v and the fact (N.4.9) that nqμ = σ to obtain, Jz = σ ( Ez + ωcτ Er) 1 1+(ωcτ)2 ωc ≡ (qB/m) Jr = σ (Er - ωcτ Ez) 1 1+(ωcτ)2 B = B θ^ Jθ = σEθ . σ = (nq2τ/m) (N.7.8) Since the radial current at the su rface "has nowhere to go" we set J r = 0 just as we set J y = 0 in the Hall effect analysis of Section N.5. One then finds Er = ωcτ Ez ( N . 7 . 9 ) where E r is a radial Hall field. Insertion of (N.7 .9) into the first line of (N.7.8) gives Jz = σ ( Ez + ωcτ [ωcτ Ez]) 1 1+(ωcτ)2 = σEz ( N . 7 . 1 0 ) Appendix N : Magnetic Ohm's Law and the Radial Hall Effect 427 and then our current components are J z = σEz Jr = 0 Jθ = 0 ( N . 7 . 1 1 ) where in the last line we have applied our ansatz that E θ = 0. Our earlier assumption that the regular Ohm's Law applies in the z direction is now self-consistently born out. The radial Hall field from (N.7.9) is Er(r) = ωc(r)τ Ez = (qB(r)/m) τ Ez = (qτ/m) B(r) E z = μ B(r) Ez // (N.1.7) for μ = μ * μ0I 2πa (r/a) * Jz σ // (N.7.4) for B(r) and (N.7.11) for E z = μ σ * μ0I 2πa (r/a) * I πa2 = 1 nq * μ0I2 2π2a3 (r/a) // (N.1.9) for μ/σ = μ0I2 2π2a3nq (r/a) ≡ Es (r/a) (N.7.12) where E s ≡ Er(a) = μ0I2 2π2a3nq volts/m . // E s < 0 since q = -|e| (N.7.13) The three electric field components are then E z = I / (π a2) Er = Es (r/a) Eθ = 0 . ( N . 7 . 1 4 ) We may then compute div E, div E = r-1∂r(rEr) + r-1∂θEθ + ∂zEz = r-1∂r(rEr) = r -1∂r(r[Es(r/a)]) = (E s/a) r-1∂r(r2) = (Es/a) r-12r = (2E s/a) (N.7.15) Since div E = ρ/ε0 , we conclude that there must be a cons tant free charge density inside the wire, ρ = ε0(2Es/ a ) . ( N . 7 . 1 6 ) In a slice of the round wire of length dz, the total internal charge is Q = ρ * (area) * dz = ρ πa 2 dz = ε0(2Es/a) πa2 dz = ε0(2πaEs)dz . (N.7.17) Appendix N : Magnetic Ohm's Law and the Radial Hall Effect 428 Since this charge had to come from somewhere, we c onclude that the outer surface of the wire slice has charge - Q and surface charge density n s ns = -Q/(2πadz) = - ε0(2πaEs)dz / (2πadz) = -ε 0Es , (N.7.18) a result one could also obtain from a gaussian box at the surface. Outside the wire, each charge density acts as a line charge at the wire center and they cancel out, so there is no external Hall field. There exists a Hall voltage between the wi re surface and the wire's center line, VH = V(a) - V(0) = ∫0 a dV dr dr = – ∫0 a Er(r) dr = – (E s/a) ∫0 a r dr = -(a/2)E s = -(a/2) μ0I2 2π2a3nq = - 1 nq * μ0I 2πa * I 2πa = – 1 nq * Bθ(a) * I 2πa (N.7.19) so VH = – RH [Bθ(a)/2π] I/ a R H = 1 nq ( N . 7 . 2 0 ) which we compare to the normal Hall effect result (N.2.5) V H = – RH Bz I / T . // the Hall voltage (N.2.5) The R H is the same in both geometries, but the thickness T is replaced by radius a, and the uniform Hall B field is replaced by B θ(a)/2π. We make this arbitrary partitioning of the factors since radius a seems the distance that most corresponds to thickness T of the normal Hall effect. It is certainly unclear how one would measure this radial Hall voltage, since it is rather difficult to place one of the voltmeter probes on the center line of a round copper wire, but doubtless this could be managed in some manner. Radial Hall Effect Hypothetical Experiment We have shown in (N.7.19) that V H = -(a/2)E s = -(a/2) μ0I2 2π2a3nq = - μ0I2 4π2a2nq = - μ0 4π2nq (I/a)2 . (N.7.21) We would like to maximize I/a in order to maximize V H, but we don't want our wire to melt. According to http://www.powerstream.com/wire-fusing-currents.htm , a fairly large I/a ratio of 45 (SI) is provided by an AWG #16 copper wire having a diameter d = 1.29 mm and a fusing current of 117 amps, so we shall run this lab experiment optimistically with I = 100 amps. What voltage V H might one observe? We have Maple evaluate these quantities: Es = μ0I2 2π2a3nq volts/m VH = -(a/2)E s volts Appendix N : Magnetic Ohm's Law and the Radial Hall Effect 429 vz = Jz/nq = (I / πa2) (1/nq) = I/(π a2nq) m/sec The results are then E r(a-ε) = Es = - 0.17 mV/m VH = 56 nV vx = - 1 8 m m / s e c ( N . 7 . 2 2 ) which can be compared with the results of our "reg ular" Hall effect experiment shown in (N.2.9). The Hall field is about 10x smaller, the Hall voltage about 350x smaller, and the drift velocity 45x larger. The Hall field just below the surface is E r(a) = Es = - 174 μV/m and decreases linearly to 0 at the wire center. Just outside the surface the field is zero since it is cancelled by the surface charge. The internal charge density ρ and the surface charge density n are then, The internal constant negative charge density ρ is very small and represents an excess of about 1 electron for every 10 21 conduction electrons. The positive surface charge n s is also tiny, being a deficiency of only 10,000 electrons per square meter. One reason the radial Hall effect is small is that the self-created B field is relatively small. On the right above Maple shows our lab example field is B θ(a) = .03T = 300 gauss, whereas in the Section N.2 the external B field was assumed to be 0.5 T = 5000 gauss. So why were we allowed to ignor e the self-generated B field in the regular Hall effect of Fig N.1? Presumably the "radial" Hall effect due to the (not shown) self-generated B field will create an internal and surface charge distribution pattern (and an internal Hall field) in Fig N.1 that is mirror-symmetric in Appendix N : Magnetic Ohm's Law and the Radial Hall Effect 430 the y = 0 plane. Thus, the regular Hall field E y gets equal and opposite radi al Hall effect contributions above and below this plane and is therefore not affected by superposing the two problems. Reader Exercise : Calculate the "radial Hall effect" for a rectangular wire like that in Fig N.1 Conclusions In the above analysis, we made certain assumptions (E θ = 0, Jz = σEz, and B = Bθ^ ) in seeking a solution for the E and B fields of an isolated, axially symmet ric infinite round wire carrying static current I. We found a solution which satisfies all four Maxwell eq uations, and since solutions are unique, that is the solution to the problem. The characteristics of this solution are: 1. There exists no radial or azimuthal current densities inside the wire, J r = Jθ = 0. The only current density is J z . 2. This current density J z is uniform over the wire cross section, so J z = I/(π a2). 3. The regular Ohm's Law applies to J z, so that J z = σ Ez. 4. The magnetic field inside the wire is given by B(r) = μ0I 2πa (r/a) θ^ . 5. In order to balance internal radial Lorentz defl ections of the current-carrying electrons, a very small internal radial E r Hall field exists inside the wire which is directed toward the center line and has the form Er(r) = Es (r/a) where E s = - μ0I2 2π2a3n|e| . (N.7.12) 6. Associated with this radial Hall field is a very sm all, negative, constant free charge distribution inside the wire which is given by ρ = ε 0(2Es/a) Q = ρ πa2 dz = ε0(2πaEs)dz 7. This fact contradicts (but in a very small way) the claim of Section 3.1 that there can be no free charge inside a conductor. That section did not incl ude the possible effect of magnetic fields. 8. This negative volume charge is extracted fro m the wire surface which then has a positive surface charge which is equal and opposite to Q shown above. Observed from outside the wire, the electric fields of these two charge distributions exactly cancel , resulting in no external radial E field. 9. We refer to the last items 5,6,7,8 above as "the radial Hall effect", for want of a better term. 10. It is hard to imagine how would might measure this effect. Appendix N : Magnetic Ohm's Law and the Radial Hall Effect 431 N.8 Magnetic Ohm's Law for Arbitrary B In Section N.4 we develo ped a Magnetic Ohm's Law for B = B z^ . We now allow B(x) to point in a general direction with components B 1, B2 and B3 and we consider first a DC static situation. Equation (N.1.5) then says, with κ = 1/μ, κ v = E + v x B where κ ≡ 1/μ // μ = (qτ/m) from (N.1.7) (N.8.1) or κv 1 = E1 + v2B3 - v3B2 κv2 = E2 + v3B1 - v1B3 κv3 = E3 + v1B2 - v2B1 . ( N . 8 . 2 ) Notice that dim( κ) = dim(B) = Tesla. Maple solves this equation for the velocity components v i : From (N.1.8) and (N.1.9) we know that J = nq v = (σ /μ)v = κσ v . Extracting the v i from the above Maple solution and multiplying by κσ we get ( N . 8 . 3 ) Appendix N : Magnetic Ohm's Law and the Radial Hall Effect 432 which is our new and very complicated tensor Magnetic Ohm's Law in the presence of an arbitrary E and B field. That is to say, we have J = Σ E where Σ is a 3x3 matrix which is a function of the B i. If we could ignore the three B i components (set them to zero in the above equations), the equations would reduce to the regular Ohm's Law J i = σEi. This is in effect the case if B i << κ for all three components of B. So a condition for the tensor Ohm's Law reduci ng to the regular Ohm's law is this: Bi << κ κ = 1/μ μ = (qτ/m) κ = (m/qτ) (N.8.4) so we need then B i << (m/qτ) . ( N . 8 . 5 ) For copper, we compute κ = m/qτ using numbers from Section N.1, Our conclusion is that "regular Ohm's Law" is applicable as long as B i << 569 Tesla. Even the largest practical B fields are far below this number. From wiki: Appendix N : Magnetic Ohm's Law and the Radial Hall Effect 433 So even the Large Hadron Collider designers and frog levitators can use regular Ohm's Law (along with the writer of Chapter 1 and Appendix D of this document). So why do we need to use the Magnetic Ohm's La w when dealing with the Hall Effect which has a relatively small B field? Recall (N.4.10), J x = σ (Ex + ωcτ Ey) 1 1+(ωcτ)2 ωc ≡ (qB/m) Jy = σ (Ey - ωcτ Ex) 1 1+(ωcτ)2 B = B z^ Jz = σEz . σ = (nq2τ/m) (N.4.10) In the Hall experiment of Fig N.1 we must have J y = 0 and that means we cannot ignore the second term in the Jy expression above, even though it is much smaller than the first term. We get E y = ωcτ Ex as in (N.5.3) which is the tiny delicate Hall field. One can repeat the above analysis to get a te nsor Magnetic Ohm's Law for a monochromatic AC situation by replacing κ → κ [ 1 + jωτ ] in (N.8.1), based on (N.1.5). As shown at the end of Section N.1, for copper we have ωτ << 1 for f << 16,000 GHz, so this κ replacement has a miniscule effect and our conclusions above still apply. Appendix O : Plotting Field Lines 434 Appendix O: How to plot 2D magnetic field lines Maple 18 and earlier versions can plot field lines (flow lines) given a function and a starting point using a certain vector calculus library package. Here we review the theory of such plots and show how the plots can be made directly. The methods given here can be generalized to make 3D plots. (a) Statement of the Problem One is given two functions Hx(x,y) and H y(x,y) which describe a 2D vector field H(x,y). This field can be directly plotted in Maple in terms of little arrows as shown for example in Fig C.2 (code shown there) using the Maple fieldplot command (this is for a recta ngular conductor with a uniform current density), Fig O.1 But we want field lines, not field arrows . One can vaguely deduce the field lines from the above picture, but we want a precise plot. (b) The Brute Force Method To track a field line, one can write a s mall spatial displacement d r in the direction of H, dr = ds H(r) . ( O . 1 ) The field line plotting code is then (pseudo Maple syntax) ds = .01 // some small number relative to the problem at hand r[1] = r 1 // pick some starting point of interest for a field line for n from 1 to 100 do d r = ds * H(r[n]) // compute a small displacement in the direction of H r [ n+1] = r[n] + d r // update position for use in next iteration od plot the list of points r[n] // this is then a field line (lis tplot, pointplot, etc) (O.2) One might gussy up the code to prevent wasted computation in locations where H is very small, perhaps computing H^ = H / |H| in each iteration then doing d r = ds * H^ . A different kind of improvement would be to use some kind of quadratic Simpson's Rule affair. Appendix O : Plotting Field Lines 435 The code above works fine, but error can build up for any finite ds. When a field line is a closed curve, the error can become visibl e where the line returns to its star ting point, as in the drawing below which shows some brute-force-method H fields lines corresponding to Fig O.1 above, Fig O.2 The field lines may seem a little surprising given the look of Fig O.1, but here is a superposition with the rectangles lined up, Fig O.3 (c) The ODE Method We outline now an alternate (and doubt less well-known) method of plotting field lines. We imagine that the description of our field H can be described by a pair of para metric equations (not yet known), x = X(s) y = Y ( s ) ( O . 3 ) where s is a real parameter. It follows that dx = (dX/ds)ds d r = (dX/ds)ds x^ + (dY/ds)ds y^ d y = ( d Y / d s ) d s . ( O . 4 ) We alter (O.1) by adding a "speed function" α (s) just because this might simplify calculations later. This function α is an arbitrary positive-definite function of parameter s. So, Appendix O : Plotting Field Lines 436 dr = α(s) ds H . ( O . 5 ) Then (O.4) and (O.5) give α(s) ds H = (dX/ds)ds x^ + (dY/ds)ds y^ α(s) ds [H x(x,y) x^ + Hy(x,y) y^ ] = (dX/ds)ds x^ + (dY/ds)ds y^ α(s) Hx(x,y) = (dX/ds) α(s) Hy(x,y) = (dY/ds) // component equations (dX/ds) = α(s) H x(X(s),Y(s)) (dY/ds) = α(s) Hy(X(s),Y(s)) . // using (O.3) (O.6) This is a pair of coupled, non-linear, first order differential equations. Conveniently, Maple knows how to numerically (and quickly) solve such a set of e quations using its dsolve command (NDSolve in Mathematica). Given the solutions X(s) and Y(s), it is th en a simple matter to plot the field lines. This is done in the following example. Example: Magnetic field lines for a two-cylinder transmission line In this exam ple we assume that th e current density in each conductor is uniform over the conductor cross section. This assumption is incorrect for a properly terminated transmission line as shown in Section 6.5, but is valid at DC and low ω for a finite-length pair of parallel wires perhaps shorted at one end to form a closed circuit. Nevertheless, we make the uniform current density approximation for a transmission line just to have a simple plotting example. Our first task is to derive expressions for the magnetic field components H x and Hy. Consider this drawing of the transmission line cross section (radii are a 1 and a2, center separation b) : Fig O.4 Current I flows into the plane of paper for the left c onductor, and out of the plane for the right. We must do a vector addition of the two magnetic fields, Appendix O : Plotting Field Lines 437 H = H1(r1) θ^1 + H2(r2) θ^2 ( O . 7 ) where θ^ 1 = -sinθ 1 x^ + cosθ1y^ θ^2 = -sinθ 2 x^ + cosθ2y^ . ( O . 8 ) From Ampere's Law for each conductor, as shown in (B.4.1) and (B.4.2), the field magnitudes are H1(r1) = (I/2π)[ θ(r1>a1) (1/r1) + θ(a1>r1) (r1/a12)] H2(r2) = -(I/2π)[ θ(r2>a2) (1/r2) + θ(a2>r2) (r2/a22) ] ( O . 9 ) where θ(x>y) = H(x-y), the Heaviside step func tion. We then have from (O.7), H = H1(r1) [-sinθ1 x^ + cosθ1y^] + H2(r2)[ -sinθ2 x^ + cosθ2y^] = [ - s i n θ1H1(r1) - sinθ2 H2(r2)] x^ + [cosθ1 H1(r1) + cosθ2 H2(r2)] y^ . But cosθ 1 = (x/r1) sin θ1 = (y/r1) cosθ2 = ((x-b)/r 2) sinθ2 = (y/r2) ( O . 1 0 ) so H = [ - (y/r 1)H1(r1) - (y/r2) H2(r2)] x^ + [(x/r 1) H1(r1) + ((x-b)/r 2) H2(r2)] y^ and the magnetic component fields are then H x = - (y/r1) H1(r1) - (y/r2) H2(r2) r 12 = x2 + y2 Hy = (x/r1) H1(r1) + ((x-b)/r 2) H2(r2) r 22 = (x-b)2 + y2 . (O.11) We now enter the expressions (O.11) and (O.9) for th e field components into Maple, setting the current arbitrarily to I = 2 π units. Both conductor radii are set to 0. 5 unit with center separation 1.25 units : The conventional Maple "field plot" can then be done this way : Appendix O : Plotting Field Lines 438 where the two red circles show the conductor surfaces, F i g O . 5 Again, we get a vague feel for what the field lines might look like. We now compute these field lines using the ODE method. So after the first block of code shown above we add instead the following: Appendix O : Plotting Field Lines 439 The two unapply commands formally make Hx_ and Hy_ functions of variables x and y. The two equation lines define ODE's eq1 and eq2 which are none other than (O.6) with α(s) = 1. We decide to plot Ncurves = 16 field lines indexed by J. The Maple dsolve command numeri cally solves the ODE's with x 0 = J*b/(Ncurves+1) and y 0 = 0 as the starting point for the curve J. Maple returns its solution as two numerically interpolated functions X(s) and Y(s) which are just those functions we assumed we had in (O.3). Special code finds an appropriate range for parameter s so curves just close on themselves, or get truncated if they go beyond a set range. Finally, the odeplot command plots the parametric f unctions X(s) and Y(s) to create the field lines in certain display data structures called p[J] for J = 1 to 16. The PLOT command makes the rectangle in p2 and finally the display command shows the results. Even on an ancient PC, this code runs in about 15 seconds. Here is the resulting plot where we have made the conductor perimeters black and the field lines red: Appendix O : Plotting Field Lines 440 F i g O . 6 Outside both conductors, the magnetic field is the same as it would be for conductors of a tiny radius, as the reader can verify by staring at (O.9). Here is the same plot with a 1 = a2 = .01 : F i g O . 7 Reader Exercise: Use H(x,y) as stated in (C.4.7), with F in (C .4.6), to plot field lines using the ODE method. Compare with the brute force method results shown in Fig O.2. Appendix O : Plotting Field Lines 441 (d) The Analytic Method The reader may notice a striking similarity between th e last plot above and Fig 6.2 which displays some Circles of Apollonius. The magnetic field lines of two pa rallel thin wires are indeed such circles, and this can be shown using the following third method of plotting field lines. From (O.5) that dr = α(s) ds H we may write dy = α(s) ds H y dx = α(s) ds Hx ( O . 1 2 ) so dy dx = Hy/Hx // right side is ratio "rat" in the code below (O.13) or Hx(x,y) dy dx = Hy( x , y ) . ( O . 1 4 ) This is a first-order non-linear ODE which Maple (or the reader) may be ab le to solve analytically for the solution y(x) which is then an analytic expression for the field line. For two thin wires, here is Maple's analytic solution using the dsolve command in its default analytic mode, Renaming the constant _C1 to be "c", and squa ring the solutions shown above, one finds that Appendix O : Plotting Field Lines 442 (x-x c)2 + y2 = r2 xc = b/4+c r2 = c2 - bc/2 - 3(b/4)2 (O.15) where recall that b is the separation of the two thin wires. Thus, the field lines are in fact circles with centers on the x axis. Reader Exercise : 1. Using the data presented in Bipolar Coordinates and the Two-Cylinder Capacitor , show that the set of circles found above are Apollonian circles with these Apollonian parameters, a = b/2 ξ = ch -1[(b/4+c)/( c2 - bc/2 - 3(b/4)2) ] ( O . 1 6 ) 2. Why might one expect Apollonian Circles for th e B field lines in this magnetostatics problem, knowing that such circles also describe the potential contours of the electrostatics problem of two cylinders? [ Hint : See (5.3.10) and (5.3.11) with β2 = k2 and (3.7.19) concerning the relation between A z and the B field lines.] Appendix P : Eddy Currents and the Proximity Effect 443 Appendix P: Eddy Currents and the Proximity Effect The Maxwell curl E equation (1.1.2) and its integral form are, curl E = - ∂tB ⇔ ∫{C E • ds = -∂t[∫S B • dS] . (1.1.36) The integral form is often written as Eemf = -∂t[magnetic flux] and one says that a changing magnetic flux through a loop induces a voltage Eemf (an "electro motive force") in that loop which then drives a current around the loop if the loop lies in a conducting medium. This is Farada y's Law of Induction and the loop of interest is usually a thin wire or coil of such wires inside, say, an electric generator. The wire or coil of wires is attached to some R load and some current I flows through the loop and load. There is Ohmic loss I2Rloop in the generating loop(s), but if R load >> Rloop this loss is minimal in the context of the generator. When the loop lies inside an open conducting medium, things become more complicated and the currents which are then driven around mathematical l oops in that medium are called eddy currents. The word eddy suggests the way water swirls around in a constrained environment when driven by wind or water currents (see Fig P.4 below). Just as the water flow velocity can have no normal component at a boundary (a steep river bank for example), an electri cal eddy current generally has no normal component at a boundary of the conductor. An exception to this ru le occurs if the eddy current is feeding a charge density on the outer surface of that boundary, and this exception would apply to water flow as well if a bank were shallow and could act as a temporary reservoi r of water. The analogy is not exact, but the word eddy is apt. In practical terms, eddy currents are normally seen as undesirable, as in a transformer core, since they represent Ohmic loss which results in power waste and heating of the core (R load = 0). Sometimes, however, eddy currents are useful, such as in non-destruc tive testing for internal cracks in metal parts, as noted below. Other eddy current applications include induction heating, object movement and levitation, and braking. There are whole books on the subject of eddy curre nts, and the web reveals a plethora of papers and theses on eddy current applications. However, simple analytic examples of eddy current problems are hard to find. The general theory is quite complicat ed, and we present below a simplified approach which suits are limited purposes. In this Appendix we explore the nature of eddy currents and compute these currents for some simple situations. We then show how one can interpret both the skin effect a nd the proximity effect (non-uniform current densities in nearby conductors) in terms of eddy currents. Appendix P : Eddy Currents and the Proximity Effect 444 P.1 Eddy Current Analysis Consider this drawing, F i g P . 1 An external apparatus has time-varying current density J ext flowing in some wires and creates a time- varying magnetic field Bext . We now bring in a Device Under Test (DUT) to obtain a new picture: F i g P . 2 For simplicity we assume that the DUT is non-magnetic ( μ = μ 0) and is a good conductor with conductivity σ. We therefore ignore displacement current s inside the DUT, in accordance with the discussion below (2.2.2). Permeability ε applies to the region outside the DUT. Iterative Interpretation We now imagine the analysis of the E and B fields to take place in the following iterative sense, where we alternately consider the tw o Maxwell curl equations: E (0) = 0 we assume no zeroth order electric field anywhere curl Bext = 0 because Jext = 0 outside the external apparatus Appendix P : Eddy Currents and the Proximity Effect 445 curl E(1) = -jω Bext time-changing Bext creates E(1) everywhere (Faraday) E = E(0) + E(1) = E(1) which is superposed onto the existing E(0) = 0 Jeddy(1) = σ E(1) E(1) inside the DUT creates Jeddy(1) in the DUT curl Beddy(1) = μJeddy(1) Jeddy(1) in the DUT creates Beddy(1) in the DUT (Ampere) curl Beddy(1) = jωε E(1) and Beddy(1) also exists outside the DUT where Jeddy = 0 . B = Bext + Beddy(1) This new B eddy(1) is superposed onto the original Bext curl E (2) = -jω Beddy(1) Beddy(1) in turn results in a further adjustment to E (Faraday) E = E(1) + E(2) which is then superposed onto the existing E Jeddy(2) = σE(2) This new adjustment to E causes an adjustment in Jeddy curl Beddy(2) = μJeddy(2) This in turn causes an adjustment to Beddy in the DUT curl Beddy(2) = jωε E(2) and outside the DUT B1 = Bext + Beddy(1) + Beddy(2) which is then superposed onto the existing B field. a n d s o o n . ( P . 1 . 1 ) Eventually we arrive at this situation inside the DUT curl ( E (1) + E(2) + ....) = -jω (Bext + Beddy(1) + Beddy(2) + ...) curl ( B ext + Beddy(1) + Beddy(2) + ...) = μσ (E(1) + E(2) + ....) = μ (Jeddy(1) + Jeddy (2) + ....) (P.1.2) which, when all is said and done, is just an iterative in terpretation of Maxwell's curl equations inside the DUT, curl E = -jω B curl B = μJ = μ(σE) ( P . 1 . 3 ) B = B ext + Beddy(1) + Beddy(2) + ... E = E(1) + E(2) + ... J = Jeddy(1) + Jeddy (2) + .... (P.1.4) One hidden assumption above is that J ext is not affected by the eddy currents and their fields, and we imagine this is implemented by some kind of curre nt source control in the external apparatus. Writing the solution of Maxwell's equations itera tively of course does not resolve the inherent complexity of the problem. For example, in the step above where we imagine computing Beddy(1), we have to compute Beddy(1) inside the DUT using curl Beddy(1) = μJeddy(1) and then we have to compute Beddy(1) outside the DUT using curl Beddy(1) = jωε E(1), and then we have to match the values of Beddy(1) on the DUT boundary. This is a full-blown boundary problem that requires much effort to solve for a general DUT and is further complicated if the DUT is made of magnetic material. Appendix P : Eddy Currents and the Proximity Effect 446 Small ω Looking at the above series of iterative steps, it would appear that if ω is in some sense "small", the series shown above for B, E and J are highly convergent and can be well approximated by the first one or two terms. In this situation, we have in essence a perturbation theory solution where ω is the smallness parameter. For very low frequencies, the key st eps in the above iterative sequence are these: curl E(1) = -jω Bext Jeddy(1) = σ E(1) (P.1.5) which we combine to get curl J eddy(1) = -jωσ Bext ( P . 1 . 6 ) and this will be the basis for the quantitative calculations of our first two examples below. Since ω is small, Jeddy(1) is small. The next iterative step curl Beddy(1) = μJeddy(1) ( P . 1 . 7 ) then results in a small B eddy(1), and then | B eddy(1)| << | Bext| . ( P . 1 . 8 ) Although we present no formal proof, it seems likely that our simple perturbative eddy current analysis can only be viable at a frequency low enough that the skin depth δ is large compared to the dimensions of the DUT. Larger ω If ω is not small, we can still write (P.1.3) as curl E = -jω (B ext + Beddy) curl (Bext + Beddy) = μ Jeddy = μσE ( P . 1 . 9 ) but the perturbation series interpretations of B eddy, Jeddy and E are not meaningful since they (probably) don't converge. In this case, we have a single monolithic problem that must be solved all at once by some method other than our simple iterative eddy current anal ysis starting with (P.1.6). What happens at higher frequencies is this: the external field B ext still creates eddy currents in th e DUT, but these currents in turn generate Beddy fields which are large enough that they significantly alter Bext within the DUT and one must then deal with B ≡ Bext + Beddy as the true field which is causing those eddy currents. In this case, one can combine the two Maxwell curl equations into a wave / Helmholtz equation as we have done in (1.2.2) and later in (1.5.27) and (1.5.32), and then one must solve that Helmholtz equation subject to appropriate boundary c onditions, both inside and outside the DUT. In effect we did this for an isolated round wire in Chapter 2 and the solution there involved extremely complicated E and B fields (recall the Kelvin functions) exhibiting sk in effect and a rapidly winding phase as shown in Figure 2.9. Appendix P : Eddy Currents and the Proximity Effect 447 Despite the difficulty of the solution for larger ω, we know a solution exists, and we can make qualitative observations about that solution b ased on the results of our simple low- ω example solutions. We shall do this below to provide an eddy current in terpretation of both the skin effect in a round wire, and the proximity effect in a transmission line. ECT Application In typical Eddy Current Testing (ECT) systems, the frequency used might range from 10Hz to 1500 Hz. The idea of an ECT system is to try to detect Beddy using a sensitive Hall Effect or SQUID device, and take note of the field pattern produced by a DUT which is "known good" (has no internal cracks in the metal). An internal crack in a bad DUT will alter Jeddy in some way, which in turn causes an alteration in Beddy which can hopefully be detected. Due to the skin depth penetration issue, the useful depth of such non-destructive testing systems might be up to 15 mm (ballpark). Higher ω generates a larger signal, gives more accuracy on the defect size and location, but penetration depth is less, so there is always a tradeoff. Often scans at different ω values are optimal for different dept hs of the defect. ECT is a subject of much current interest and many pape rs have been and are being written. P.2 Eddy currents in a thin round plate in a uniform B field To reduce sym bol clutter, in this section we use the following notation in relation to Section P.1 : B ≡ Bext J ≡ Jeddy(1) (P.2.1) A thin round plate of radius a and thickness h lies centered in the z = 0 plane of a cylindrical coordinate system. This plate is the Device Under Test (DUT) fo r this problem. An unseen external apparatus creates a time-varying and spatially uniform magnetic field B = B z ^ perpendicular to the plate. The problem is to compute the electric field E in the plate, and the corresponding eddy currents J = σ E. The region surrounding the plate is assumed non-conducting, perhaps it is air. Faraday's Law and symmetry imply circular closed elect ric field lines in the plate. We are perhaps more used to magnetic field lines being closed since div B = 0, but here we have cl osed electric field lines since div E = 0 inside the plate (since there is no free charge inside the plate, see Section 3.1). We assume sufficiently low ω so skin depth δ >> h ( recall δ ≡ 2/(ωμσ) ) so the E field lines are uniform in the z direction of the plate thickness. This is the implication of the word "thin" in discussing a "thin plate". In this drawing, the plate is gray, and some of the circular closed E field lines are shown in red, Appendix P : Eddy Currents and the Proximity Effect 448 Fig P.3 The magnitude of the E field is constant on each circle due to symmetry. The field lines are drawn clockwise since we know by Lenz's Law that the associated eddy currents pr oduce a B field opposed to the applied B field. Since ω is small, we may use our first pertur bation theory expansion term (P.1.6), curl J eddy(1) = -jωσ Bext . (P.1.6) Setting J ≡ J eddy(1), B ≡ Bext, J = σE, and jω → ∂t, this says, curl E = - ∂tB ⇔ ∫{C E • ds = -∂t[∫S B • dS] (P.2.2) which we recognize as the Maxwell curl E equation and its integral form as shown in (1.1.36). Section P.1 has provided a context and an interpretation of the symbols B, E and J = σE appearing in (P.2.2). Consider a circle at radius r. We then have from the right equation of (P.2.2), E θ * 2πr = - B• πr2 => E θ(r) = - B• πr2/2πr = (-B• /2) r . (P.2.3) The E field is linear in r, and is reminiscent of the result for the H field inside a round wire which carries a DC current I, H * 2 πr = I πr 2/πa2 =< H(r) = I r2/a2 / 2πr = (I/2πa2)r . (C.3.9) Notice that the E field-line circles continue outside the plate and under and over it, but there will only be current in the plate. Here then is our result for E and J inside the plate: E(r) = Eθ(r) θ^ Eθ(r) = (-B• /2) r J θ(r) = σ (-B• /2) r . (P.2.4) We digress momentarily to study the heat loss generated in the plate. Consider a thin cylindrical shell of height dz and radius r and thickness dr. Think of this as a circular wire of rectangular cross section dA = dzdr. The current in this wire is given by, Appendix P : Eddy Currents and the Proximity Effect 449 dI = J θ dA = J θ d z d r . ( P . 2 . 5 ) The resistance of the wire is dR = ρ L/dA = ρ 2πr/(dzdr) where ρ = 1/σ. The power burned from P = I2R is dP = (dI) 2dR = [J θ dzdr]2 * ρ 2πr/(dzdr) = J θ2 [dzdr] * ρ 2πr = J θ2 2πrρ dr dz = [ σ (-B• z /2) r]2 2πrρ dr dz = ( π/2)σ B• z2 r3 drdz . (P.2.6) Now integrate over the plate thickness h to replace dz by h. Then integrate r from 0 to a to get P = (π /2)σ B• z2 h(a4/4) = (π/8)σ B• z2a4h . (P.2.7) Dimensions: RHS = [ohm-1m-1] sec-2 [volt-sec/m2]2 m5 = ohm-1sec-2 [volt-sec]2 = volt2/ohm = watts The total current going around the plate as observed through any azimuthal slice θ = θ 1 is, I = h ∫0 a dr Jθ(r) = h σ (-B• z /2) ∫0 a rdr = h σ (-B• z /4)a2 = - (1/4) h σa2B• z . (P.2.8) To summarize our conclusions for eddy currents in the thin round plate of radius a, thickness h and conductivity σ , J θ(r) = - (1/2) σB• r (P.2.4) // the eddy current density I = - (1/4) σha2B• (P.2.8) P = (π /8) σha4 B•2 (P.2.7) ( P . 2 . 9 ) These results are in agreement with equations (36), (37) and (38) of Siakavellas. Since B• → jωB, the power loss is proportional to the square of the frequency of the external B field, and it is proportional to the conductivity of the plate and its thickness. [Siakavellas also treats thin plates with polygonal boundaries.] It is a simple matter now to plot the ed dy current vector inside the plate: J x = Jθ θ^•x^ = -Jθ sinθ = -Jθ (y/r) = + (1/2) B•σ y ≡ ky θ(r<a) k = (1/2) B•σ Jy = Jθ θ^•y^ = Jθ cosθ = Jθ (x/r) = -(1/2) B•σ x ≡ - kx θ(r<a) . (P.2.10) With k = 1 and a = 1 Maple produces the following plot of the eddy currents inside the plate: Appendix P : Eddy Currents and the Proximity Effect 450 Fig P.4 With k = 1 we have assumed that B• > 0 (out of plane of paper), and according to Lenz's Law, the eddy current creates a B field whose flux cancels some of th e applied B flux, hence the clockwise direction. Reader Exercise: Notice that the E eddy current arrows are largest near th e edge of the plate according to (P.2.4) which says E θ(r) = constant * r. Is it correct to interpre t this as a 2D skin effect of the type encountered in Chapter 2? In terms of the external B field penetrating the disk, since the disk is thin we have assumed no skin effect in the z dimension and that B field penetrates fully and δ >> h. Reader Exercise: The EMF due to Faraday induction drives J eddy in a tangential direction as Fig P.4 shows. Each conduction electron appears to maintain its radius from the disk center. Interpret this behavior in terms of the Lorentz force acting on the electron. Is there a "radial Hall effect" present in this problem? (see Appendix N) Appendix P : Eddy Currents and the Proximity Effect 451 P.3 Eddy currents in a thin round plate in a non- uniform B field This example is the same as that of the previous section, except the B field is no longer spatially uniform and for simplicity we assume it has the following simple linear form, B(x) = B 0 - α x . α > 0 // B(x) larger for x < 0 (on the left) (P.3.1) This is then a simple case where B ext(x) = B(x ) has a gradient over the plate. Even with this simple form, the problem is consid erably more complicated that the previous problem since we can no longer make use of azimuthal symme try. First consider (P.2.2) now in the frequency domain, where as before B ≡ Bext and E = Jeddy/σ. (This equation is really (P.1.6) of Section P.1. ) curl E = - jω B . (P.2.2) (P.3.2) In cylindrical coordinates this says r^ [ r-1∂θEz - ∂zEθ] + θ^ [∂zEr - ∂rEz] + z^ [ r-1∂r(rEθ) - r-1∂θEr ] = - jω Bz^ . (P.3.3) As in the previous example, we assume the plate is very thin and ω is very small so δ >> h, so fields are constant in the z direction allowing us to replace ∂z → 0. At the same time, we assume E z = 0 since E z has no apparent source (and J z has nowhere to flow). Then the above vector curl equation boils down to this scalar equation for the z component, r -1∂r(rEθ) - r-1∂θEr = - jω [B0 - α x] or ∂ r(rEθ) - ∂θEr = - jω [B0 - α rcosθ] r ( P . 3 . 4 ) since x = rcos θ. Since there is no charge inside the plate (as before), we know div E = 0, or div E = r-1∂r(rEr) + r-1∂θEθ + ∂zEz = 0 . ( P . 3 . 5 ) Again we set E z= 0 (or ∂ z→0) which kills off the last term. Then (P.3.3) and (P.3.4) may be written ∂ r(rEθ) - ∂θEr = - jω [B0 - α rcosθ]r ∂ r(rEr) + ∂θEθ = 0 . (P.3.6) These equations form a system of coupled fi rst-order linear PDE's in variables r and θ for functions Er(r,θ) and E θ(r,θ). There is no z argument since we assumed ∂z → 0 above, so basically we have a 2D problem in polar coordinates. The first equation is inhomogeneous (has a driving term) while the second is homogeneous. We wish to emphasize how the addition of a simple linear external B field variation has converted the trivial problem of Section P.2 to a non-trivial probl em involving coupled par tial differential equations. There are of course associated boundary conditions, such as E r(a,θ) = 0 since J r can have no normal Appendix P : Eddy Currents and the Proximity Effect 452 component at the rim of the plate. This complexity is typical of "eddy current problems". Below we shall solve this problem using a potential method, and one can then show that the solutions so obtained do in fact satisfy (P.3.6). (a) The stream function method in Cartesian Coordinates In our treat ment of transmission lines in Chapter 4, we used the fact that div B = 0 to describe the magnetic field in terms of a magnetic vector potential A, where B = curl A (since div curl A = 0 for any A). Under suitable conditions, it was then possible to ignore the "transverse" components of A and deal only with the z component A z. This then replaced the complexity of three fields B i with one field A z. In our current context of dealing with electric fields inside a conductor (where ρ = 0) we have div E = 0 and therefore div J = 0 since J = σE. We can then describe the current J in terms of a current vector potential T, where J = curl T (since div curl T = 0 for any T). For a thin plate at low frequency ω, we will argue that the transverse components of T may be neglected, and then the complexity of three fields J i is replaced by one field T z. This field is known in incompressible fluid dynamics as a stream function where J = ne v is essentially the fluid flow velocity field v. Here then is the stream function method presented in Cartesian coordinates. First, J = curl T = x^ (∂ yTz - ∂zTy) + y^ (∂zTx - ∂xTz) + z^ (∂xTy - ∂yTx) Jx = ∂yTz - ∂zTy Jy = ∂zTx - ∂xTz Jz = ∂xTy - ∂yTx . // components of the above (P.3.7) The J we have in mind is J eddy(1) appearing in (P.1.6). Using the symbols of (P.2.1), we have curl J = - jωσ B (P.1.6) ( P . 3 . 8 ) where B = Bz z^ [ = Bext] . Using a standard vector identity we find then that curl J = curl curl T = grad(div T) - ∇2T . (P.3.9) Just as we are allowed to work in the "Coulomb gauge" div A = 0 with the magnetic vector potential (see Appendix A), here we can work in the div T = 0 gauge for the current vector potential. In this gauge, we combine (P.3.8) and (P.3.9) to get ∇2T = jωσB ( P . 3 . 1 0 ) where ∇2 is the vector Laplacian operator. This equation can be compared with ∇2A = - μJ which is (1.3.5) for a magnetostatic situation where A has no time dependence. As shown in (H.1.9), equation (P.3.10) has the solution T(x) = -∫d3x' [1/4πR][jωσB] + possible homogeneous solutions R = | x-x'| (P.3.11) Appendix P : Eddy Currents and the Proximity Effect 453 where the integral is the "particul ar solution" of (P.3.10). Since B = Bext = Bz(x) z^, the particular solution is entirely in the z direction. There may be some homogeneous adder solutions which create transverse components T x and Ty, but we make the ansatz that Tx, Ty << Tz . ( P . 3 . 1 2 ) One motivation for this assumption is that then J z = ∂xTy - ∂yTx of (P.3.7) will be very small as we expect for a "thin" plate (we already assumed E z= 0 above). Bypassing a detailed analysis of this issue, we shall assume that T x = Ty = 0 and only T z is significant. Then (P.3.7) becomes J x = ∂yTz Jy = - ∂xTz Jz = 0 . ( P . 3 . 1 3 ) Furthermore, we assume that T z = Tz(x,y) with no z dependence, since th en (P.3.13) will lead to currents Jx and Jy which have no z dependence. With all these assu mptions, (P.3.10) becomes a scalar equation ∇2D2Tz(x,y) = j ωσBz( x , y ) ( P . 3 . 1 4 ) where ∇ 2D2 ≡ ∇2 - ∂z2 is the transverse component of the 3D scalar Laplacian. Equation (P.3.14) is just the 2D Poisson equation of 2D potential theory [see (A.0.1) for the normal Poisson equation in 3D ]. Many tools are available fo r solving this equation, and we shall use some of these tools below in our solution of the thin plate problem. (b) The stream function method in Cylindrical Coordinates In c ylindrical coordinates (really polar coordinates) one writes ∇2D2 Tz = r-1∂r(r∂rTz) + r-2∂θ2Tz so (P.3.14) becomes r -1∂r[r ∂rTz(r,θ)] + r-2∂θ2Tz(r,θ) = jωσ Bz(r,θ) . ( P . 3 . 1 5 ) The ansatz (P.3.12) becomes Tr, Tθ << Tz . ( P . 3 . 1 6 ) The cylindrical replacement for (P.3.7) is J = curl T = r^ [ r -1∂θTz - ∂zTθ] + θ^ [∂zTr - ∂rTz] + z^ [ r-1∂r(rTθ) - r-1∂θTr ] Appendix P : Eddy Currents and the Proximity Effect 454 Jr = r-1∂θTz - ∂zTθ Jθ = ∂zTr - ∂rTz Jz = r-1∂r(rTθ) - r-1∂θTr // components of the above (P.3.17) which, using (P.3.16), we approximate as Jr = r-1∂θTz Jθ = - ∂rTz Jz = 0 ( P . 3 . 1 8 ) (c) Using the stream function method to solve the plate problem From (P.3.1) we have B z(x) = B0 - α x = B0 - α rcosθ, so (P.3.15) states that ∂ r[r∂rTz(r,θ)] + r-1∂θ2Tz(r,θ) = jωσ [B0 - α rcosθ] r . (P.3.19) Here we have a single second-order PDE in r, θ for a single function T z(r,θ) [ the stream function]. One can compare this with the pair of coupled first-order PDE's found earlier in (P.3.6). Since the angle θ has the full range (0,2 π) we can expand the various func tions into "partial waves" as shown in (D.1.5) for a scalar function, so T z(r,θ) = ∑ m = -∞∞ Tz(r,m) ejmθ (D.1.5a) Tz(r,m) = (1/2 π) ∫-π π dθ Tz(r,θ) e-jmθ (D.1.5b) (P.3.20) where we use our usual overloaded notation for T z. Then (P.3.15) becomes, using ∂θ→ +jm, ∂r[r∂rTz(r,m)] - r-1m2Tz(r,m) = jωσ Bz(r,m) r (P.3.21) where B z(r,m) = (1/2 π) ∫-π π dθ Bz(r,θ) e-jmθ = (1/2π) ∫-π π dθ [B0 - α rcosθ] e-jmθ = B 0 (1/2π) ∫-π π dθ e-jmθ - α r (1/2π) ∫-π π dθ cosθ e-jmθ = B 0 (1/2π) ∫-π π dθ cos(mθ) - α r (1/2π) ∫-π π dθ cosθ cos(mθ) = B 0 (1/π) ∫0 π dθ cos(mθ) - α r (1/π ) ∫0 π dθ cosθ cos(mθ) Appendix P : Eddy Currents and the Proximity Effect 455 = δm,0B0 (1/π) π - α r (1/π ) δm,±1 π/2 = δm,0B0 - (α/2) r δm,±1 ( P . 3 . 2 2 ) where we use the following integral for integers m and n, ∫0 π dθ cos(mθ)cos(nθ) = ⎩⎨⎧ 0 m ≠ n π/2 m = n ≠ 0 π m = n = 0 . // Spiegel p 96 15.27 Equations (P.3.21) become ∂r(r∂rTz(r,0)) = jωσ B0 r m = 0 ( P . 3 . 2 3 ) ∂ r(r∂rTz(r,±1)) - r-1Tz(r,±1) = - jωσ (α/2) r2 m = ± 1 ( P . 3 . 2 4 ) ∂ r(r∂rTz(r,m)) - r-1m2Tz(r,m) = 0 m = other integers (P.3.25) We assume the relevant solution to (P.3.25) is T z(r,m) = 0. Maple tells us the general solutions to the first two equations, We rename the constants to write these solutions as, T z(r,0) = jω σ B0 (1/4)r2 + C1 ln(r) + C 2 T z(r,±1) = - (1/8) j ωσ (α/2) r3 + D1(r-1/r) + D 2(r+1/r) . (P.3.26) To have T z(r,0) finite at r = 0 we must have C 1 = 0. To have T z(r,±1) finite at r = 0 we must have D 1 = D2. The solution forms are then Appendix P : Eddy Currents and the Proximity Effect 456 Tz(r,0) = jωσ B0 (1/4) r2 + C2 Tz(r,±1) = - (1/8) j ωσ (α/2) r3 + 2D1 r . ( P . 3 . 2 7 ) Inserting these partial wave amplitudes into (P.3.20) gives Tz(r,θ) = Tz(r,0) + T(r,+1)ejθ + T(r,-1)e-jθ = Tz(r,0) + T(r,+1) 2 cos θ = [ j ωσ B 0 (1/4) r2 + C2] + 2 cosθ [- (1/8) jωσ (α/2) r3 + 2D1 r] = [ j ωσ B 0 (1/4) r2 + C2] - cosθ r [ (1/8) j ωσ α r2 - 4D1] . (P.3.28) At the origin point r = 0 we arbitrarily set the potential T z(0,θ) = 0 so C 2 = 0. One always has this freedom with a potential: since J = curl T, constants in T don't affect J. We then compute J r as shown in (P.3.18) to obtain Jr(r,θ) = r-1∂θTz = sinθ [(1/8) jωσ α r2 - 4D1 ] . ( P . 3 . 2 9 ) But at r = a, we must have J r(a,θ) = 0 since there can be only tangential currents at the rim of our circular plate, and this determines D 1 giving this final result for the stream function T z , T z(r,θ) = jωσ B0 (1/4) r2 - (1/8) cos θ r jωσ α (r2-a2) = j ωσ [ (1/4)B 0 r2 - (α/8) cosθ (r3-a2r ) ] . ( P . 3 . 3 0 ) The eddy current components ar e then, again from (P.3.18), Jr(r,θ) = r-1∂θTz = r-1 jωσ (α/8) sinθ (r3-a2r) = jωσ [(α/8) sinθ (r2-a2)] J θ(r,θ) = - ∂rTz = jωσ [ - (1/2)B 0 r + (α/8) cosθ (3r2-a2) ] . (P.3.31) Here then is a summary of the solution for a circular plate of radius r and conductivity σ in the presence of an external magnetic field B(x) = B 0 - α x : Tz(r,θ) = jωσ [ (1/4)B 0 r2 - (α/8) cosθ r (r2-a2) ] J r(r,θ) = jωσ [(α/8) sinθ (r2-a2)] Jθ(r,θ) = jωσ [ - (1/2)B 0 r + (α/8) cosθ (3r2-a2) ] E r(r,θ) = jω [(α/8) sinθ (r2-a2)] Eθ(r,θ) = jω [ - (1/2)B 0 r + (α/8) cosθ (3r2-a2) ] . ( P . 3 . 3 2 ) Appendix P : Eddy Currents and the Proximity Effect 457 The eddy currents J r and Jθ are proportional to ω as expected from (P.1.6). Current J r vanishes at the edge of the plate. When α = 0, we find J r(r,θ) = 0 and J θ(r,θ) = jωσ [ - (1/2)B 0 r ] which replicates our uniform-B solution (P.2.4) which was J θ(r) = σ (-B• /2) r . We have verified using Mapl e that the stream function T z(r,θ) shown in (P.3.32) satisfies (P.3.19) and that the electric fields E r(r,θ) and E θ(r,θ) satisfy (P.3.6). Finally, we have Maple plot the resulting eddy currents. We first write Jx = Jr r^•x^ + Jθ θ^•x^ = Jr cosθ – Jθ sinθ Jy = Jr r^•y^ + Jθ θ^•y^ = Jr sinθ + Jθ cosθ . ( P . 3 . 3 3 ) For plotting we set j ωσ = 1, a = 1, B 0 = 1, and α = 1. Here is the plotting code, and here is the resulting plot, Appendix P : Eddy Currents and the Proximity Effect 458 Fig P.5 which one can compare with the no-gradient plot of Fi g P.4. As expected, the eddy currents are larger on the left side because the external B field is larger there. One can imagine the E field lines being a set of distorted circles which shrink down about a point to the left of the origin. The main point of this example is to demonstrate the fact that a gradient in th e external B field results in an asymmetry in the eddy current distributio n such that the larger eddy cu rrent vectors are in the region in which the external B field is largest. We shall see below in a different geometry how this fact accounts for the so-called proximity effect in a transmission line. Reader Exercise: Use one of the methods of Appendix O to plot the E field lines for Fig P.5. P.4 Self-induced eddy currents in a round wire We now reconsider our well-studied axially symmetric radius-a round wire of Chapter 2. In this eddy current example, the "external apparatus" and the "device under test" (DUT) are one in the same! In the zeroth order of the Section P.1 perturbation theory (very low ω), the current density in the wire is Jext(x,ω) which is perfectly uniform across the wire cross section and flows in the z^ direction. This current density creates a magnetic field Bext in the θ^ direction which is obtained from Ampere's Law, 2πr Bext(r) = πr2 πa2 μ Ienc = πr2 πa2 μ Jext πa2 = πr2μJext Appendix P : Eddy Currents and the Proximity Effect 459 => B ext(r) = (1/2) μ r Jext . (P.4.1) In this problem the "external" current density J ext is generated by the round wire itself, as if it were somehow its own "external appara tus". A better notation would be Jdc since this is the ω = 0 current distribution, but we continue to use J ext to maintain contact with the Section P.1. Similarly, Bext = Bdc . If the skin depth δ is large compared to the wire radius a, we expect the perturbation eddy current analysis of Section P.1 to be viable, and we write (P.1.6) as curl J eddy ≈ - jωσ Bext where Bext = (1/2) μ r Jext θ^ = Bext(r) θ^. (P.4.2) In cylindrical coordinates one writes for an arbitrary vector field F, curl F = r^ [ r-1∂θFz - ∂zFθ] + θ^ [∂zFr - ∂rFz] + z^ [ r-1∂r(rFθ) - r-1∂θFr ] . (P.4.3) For a vector field F which is a function only of r this reduces to, curl F = θ^ [- ∂rFz] + z^ [ r-1∂r(rFθ) ] . ( P . 4 . 4 ) Thus (P.4.2) becomes these two equations, r -1∂r[r(Jeddy)θ] = 0 - ∂ r(Jeddy)z = -jωσ Bext(r) = -jωσ (1/2) μ r Jext . (P.4.5) The first equation of (P.4.5) may be written as ∂ r[r(Jeddy)θ] = 0 or [r(J eddy)θ] = C1 or (Jeddy)θ(r) = C1/r from which we must conclude that C 1 = 0 and then (J eddy)θ(r) = 0, so there is no azimuthal eddy current in the wire. The second equation of (P.4.5) may be integrated from r=0 to r=r to obtain (J eddy)z(r) - (Jeddy)z(0) = jωσ (1/2) μ Jext ∫0 r dr' r' = jωσ (1/4) μ Jext r2 . (P.4.6) Thus, the total current density in the wire obtained from eddy current analysis is in the z direction and is given by Appendix P : Eddy Currents and the Proximity Effect 460 Jz = Jext + (Jeddy)z = Jext + (Jeddy)z(0) + jωσ (1/4) μ Jext r2 = J ext [ 1 + (Jeddy)z(0) Jext + j ωμσ (1/4) r2 ] ≈ Jext [ 1 + j ωμσ (1/4) r2 ] . // since |(J eddy)z(0)| << |J ext| at low ω (P.4.7) Recall from (2.2.3) and (2.2.20) that jβ 2 = ωμσ = 2/δ2 = |β2| // j ωμσ = -β2 (P.4.8) where β is the complex Helmholtz parameter of (1 .5.1c). Therefore we have shown that J z = Jext [ 1 + j ωμσ (1/4) r2 ] = Jext [ 1 - β2(1/4) r2 ] . (P.4.9) Notice that the eddy current contribution is π/2 out of phase with J ext. Since we have assumed δ >> a, it follows that |βa| = ( 2 /δ) a = 2 (a/δ) < < 1 ( P . 4 . 1 0 ) so then |βr| << 1 and the eddy current contribution is very small, as required to use the first term in the perturbation expansion of Section P.1 as we have done. Defining κ ≡ ωσμ (1/4)r2 = (jβ2) (1/4) r2 = | β2| (1/4)r2 << 1 jκ = jωσμ (1/4)r2 = -β2(1/4) r2 ( P . 4 . 1 1 ) one finds J z = Jext [ 1 - (β2/4)r2] = Jext [ 1+jκ ] |Jz|2 = |Jext|2 (1+jκ)(1-jκ ) |Jz| = | Jext | (1+jκ)(1-jκ ) = | Jext | 1 + κ2 ≈ | Jext | [ 1 + (1/2)κ2] (P.4.12) so that |J z| | Jext | = [ 1 + (1/2) κ2] = [ 1 + (1/2) { | β2| (1/4)r2 }2] = 1 + 1 2 {(2/δ2) (1/4)r2}2 = 1 + 1 2 {(1/δ2) (1/2)r2}2 = 1 + 1 8 (r/δ)4 ( P . 4 . 1 3 ) which then exhibits a very slight skin effect a nd has the same r dependence as (2.3.10), see Fig 2.7. In Chapter 2 we found in (2.2.30) the following exact result for J z in a round wire operating at ω, Appendix P : Eddy Currents and the Proximity Effect 461 Jz(r) = I 2πa J0(βr) J1(βa) β . (2.2.30) For small ω, since |βr| << 1, one has for small arguments [ Spiegel 24.5 and 24.6 ] J0(x) ≈ 1 - x2/4 J1(x) ≈ (x/2)(1 - x2/8) ⇒ 1/J1(x) ≈ (2/x) (1 + x2/8) ≈ (2/x) (P.4.14) so J0(βr) J1(βa) ≈ (2/βa) (1-β2r2/ 4 ) ( P . 4 . 1 5 ) and then J z(r) = I 2πa [(2/βa) (1-β2r2/4)] β = I 2πa [(2/a) (1- β2r2/4)] = I πa2 [ (1-β2r2/4)] = J ext (1-β2r2/ 4 ) ( P . 4 . 1 6 ) in agreement with our eddy curre nt analysis result (P.4.9). At higher frequencies where we no longer have δ >> a, the eddy current pert urbation expansion diverges and becomes meaningless and one must instead solve th e Helmholtz equation stated at the end of Section P.1. In Chapter 2 this task was in essence carried out and the skin effect was observed. One can then interpret the skin effect by saying that the eddy curre nts cancel the DC current density in the interior of the round wire, allowing a net current to exist only at th e periphery. In other words, the skin effect is caused by eddy currents. But this is just a manner of sp eaking, and is like saying that the skin effect is "caused by Maxwell's Equations", which it is. For a moderate skin effect, we can illustrate the eddy currents in a round wire by crudely plotting them just in the central gray plane of the following drawing : Fig P.6 Theses qualitative-only plots are for some particular instant in time. We know that the phase of J varies as shown in Fig 2.8, so we attempt to illustrate only the real parts the currents: Appendix P : Eddy Currents and the Proximity Effect 462 R e { J ext} F i g P . 7 ( a ) R e { J eddy} F i g P . 7 ( b ) Re{J ext+Jeddy} = skin effect Fig P.7 (c) The closed red curves in Fig P.7 (b) represent the J eddy field lines, and these then represent the actual induced "eddies" of current. One could write Jeddy = σ Eeddy and then they are electric field lines. The lines close on themselves because they have no sources: inside the wire ρ = 0 so div Jeddy = 0 and div Eeddy = 0. Recall that a field in general does not have a constant magnitude along a field line. In the geometry of a round wire, the field lines in fact loop around at the ends of the wire. P.5 Eddy currents induced in a quiet round-wire by an external B field Uniform Bext In this example, we start with our Device Under Test (DUT) which is a straight round wire which carries no current. Some "external apparatus" creates a time-changing magnetic field Bext = Bexty^ as shown in Appendix P : Eddy Currents and the Proximity Effect 463 the figure below, where B ext(x,ω) is for the moment constant in space. At the instant in time shown, the time-domain field Bext(x,t) is increasing in the - y^ direction so that - B• ext(x,t) points in the + y^ direction, out of the plane of paper. F i g P . 8 The time-domain eddy current equation (P.1.6) ( we assume small ω) and its integral form are curl Jeddy = σ [- B• ext] ⇔ ∫{C Jeddy • ds = σ ∫S [- B• ext] • dS . (P.5.1) The integral form implies that the flux change thro ugh any math loop in the gray rectangle is positive at our time instant, so according to the right hand rule , the eddy currents in the gray rectangle have the following general shape, Fig P.9 This figure is analogous to Fig P.4 above which shows the eddy current in a thin round plate for a uniform external B field. To justify the linear variation with x (vertical), if we assume that away from the ends of the wire nothing varies with z, and if we write J eddy as J, we find that curl J = x^ (∂yJz - ∂zJy) + y^ (∂zJx - ∂xJz) + z^ (∂xJy - ∂yJx) = - σ B• exty^ or x^ (∂ yJz) + y^ ( - ∂xJz) + z^ (∂xJy - ∂yJx) = - σ B• exty^ (P.5.2) which produces the three equations Appendix P : Eddy Currents and the Proximity Effect 464 ∂xJz = σ B• ext => J z(x) = σ B• ext x // linear in x ∂yJz = 0 => J z = Jz(x) only ∂xJy - ∂yJx = 0 satisfied if J x and Jy = 0 (P.5.3) The eddy pattern in the round wire would have the same general appearance in any slice of the wire parallel to the slice shown as the gray rectangle in Fi g P.8. The current of course drops to 0 at the wire surface since we assume the wire is surrounded by an insulating medium. Conversely, in any planar slice of the wire which is perpendicular to the gray plane (and still parallel to the z axis), there are no eddy currents because any math loop in such a plane sees no flux. Non-uniform B ext Suppose now that the field B ext has a positive linear gradient in the x direction. We can write, jω Bext(x,ω) = jω[ Bext0 + αx] => B• ext(x,t) = B• ext0 + α• x (P.5.4) Then, B• ext(x,t) = B• ext(t) + α•(t) x = j ω ejωt [ Bext(0) + α(0) x ] . (P.5.5) We assume α (0) > 0 so the B field magnitude is larger at th e top of Fig P.8 than at the bottom at t = 0. Below we shall assume a time such that ejωt = -1 so then both B• ext(t) and α•(t) are negative. Then the first equation of (P.5.3) becomes, ∂xJz = σ B• ext = σ [B• ext0 + α• x ] => J z(x) = σ [B• ext0 x + (1/2) α• x2 - (1/6) α• ] or J z(x) = - σ [ | B• ext0| x + (1/2) | α• | x2 - (1/6) | α• | ] // for our time of interest (P.5.6) where we have added a constant such that ∫-1 1 dx Jz(x) = 0 for a wire of radius a = 1. J eddy = Jz is now larger in the upper half of the gray rectangle than in the lower half, and we would expect then a pattern having this general shape, Appendix P : Eddy Currents and the Proximity Effect 465 Fig P.10 The eddy currents are now larger on the side of the wire where the external field B ext is larger. In addition, one sees the eddy current in general to be la rger near the surface of the wire and small in the interior. Figure P.10 is analogous to Fig P.5 above which shows the eddy current in a thin round plate for a non-uniform external B field. P.6 Eddy currents induced in an current-c arrying wire by an external B field We start with the self-induced eddy current patte rn in a round wire shown in Fig P.7 (b), Fig P.7 (b) We then turn on a uniform external B field which in duces an additional eddy current pattern in the same round wire, as shown in Fig P.9, Fig P.9 We assume the external B field has the proper phase re lative to the current in the wire such that the patterns look as shown above. When a small amount of the lower pattern is superposed on the upper Appendix P : Eddy Currents and the Proximity Effect 466 pattern, there is partial cancellation on the top edge, and reinforcement on the lower edge, resulting in the following eddy current pattern, Fig P.11 Thus we arrive at another mechanism for the eddy curre nt to be larger on one side of a wire than on the other side. Notice that B ext has no gradient in this example, and also that we are not showing the underlying DC uniform current pattern of Fig P.7 (a). P.7 Summary of Round Wire Examples 1. The eddy currents which a current-carrying round wire induces into itself vary with radius inside the wire, but are azimuthally symmetric. These eddy curren ts are interpreted as causing the skin effect. This situation is depicted in Fig P.7 (c). 2. A quiet round wire in the presence of a spatially-unifo rm external B field will have induced eddy currents which are oppositely directed on the two sides of the wire, but the absolute value of the current is symmetric on the two sides, as in Fig P.9. 3. If this quiet wire is placed in an external B ext field which has a gradient , then the absolute value of the current density will be larger on the side of the wire where B ext is larger, as shown in Fig P.10. 4. When a current-carrying round wire is placed in a uniform external B ext field, even though that field is uniform, the absolute value of the current density is larger on one side of the wire compared to the other side, as shown in Fig P.11. 5 When a current-carrying round wire is placed in a external B ext field which has a gradient, we again expect to have a side-to-side eddy current asymmetry which is a combination of the effects of items 3 and 4 above. The asymmetry will depend on the direction of the current in the wire and on the size and polarity of B ext. P.8 Eddy currents in Transmission Lines: The Proximity Effect We consider a transm ission line composed of two round wires and focus our attention on wire #1 as our Device Under Test. If wire #2 is far away, as in a wide-spaced twin-line, then B ext created by wire #2 is roughly uniform at the location of wire #1, and we then have the current asymmetry of Case 4 above. If wire #2 is close to wire #1, then B ext created by wire #2 will have a gradient over wire #1, and then we Appendix P : Eddy Currents and the Proximity Effect 467 have the asymmetry combination Case 5 above where both effects must be considered. In the following drawing, we show in cross section two round wires bot h of which carry current I in the same direction, out of the plane of paper : F i g P . 1 2 The field B ext created by wire #2 is slightly stronger on the right side of wire #1 than on the left side, which of itself would argue for more eddy current on th e right side of wire #1. However this effect is swamped by the Case 4 effect where we have cancellation of B fields on the right side of wire #1 and addition on the left side, so the total B field is st ronger on the left side of wire #1 and thus the eddy current (and hence the total current density) is larger there, as indicated by the lighter coloration. The current asymmetry increases as the two conductors get closer together because both the B field cancellation and reinforcement are enhanced as B ext becomes larger and more comparable to the internal B field. The asymmetry also increases as ω increases, since the eddy currents increase, and at ω = 0 there is no asymmetry because there are no eddy currents. If we imagine positive charge carriers coming out of th e plane of paper in wire #1, the ones on the left side of wire #1 feel a Lorentz force q v x B pushing them to the right, while those on right side of wire #1 feel an oppositely directed force pushing them to the left. This is so because the net B in general points down on the left side of the center line of wire #1, and up on the right side (see Fig P.16 below). But the charge carriers on the right are in a smaller B field and travel at a smaller velocity v since J z = nqv is smaller (although at ω = 0 this second fact is not true). The net effect is that for any ω ≥ 0 the charge carriers in wire #1 feel an overall fo rce to the right and this force is transferred to the conductor ion lattice to maintain ρ = 0 causing the entire wire to be pushed to the right. The opposite happens inside wire #2 and the result is that wires with currents in the same direction attract each other for any ω ≥ 0. The fact that (for ω > 0) the current distribution in each wire is skewed away from the other wire is sometimes called the proximity effect , or current crowding . One effect of having a non-uniform J z distribution is that the wires have resistance larger than their DC values (see Section P.10 below). If the above two wires were two strands of a power tr ansmission line cable carrying current in the same direction, the Ohmic loss in the strands is enhanced by this crowding effect. The coloration patterns in Fig P.12 don't really illu strate the skin effect which is of course always present at any ω > 0, more strongly of course when δ < a (see Fig 6.12.). Even in power lines at 60 Hz where δ ~ 1 cm the skin effect does cause a waste of the wire interior since less current flows there. If a large round conductor is replaced by a set off smaller insulated round wires, this waste is reduced since the smaller wires each have more uniform current (see Litz wire). Appendix P : Eddy Currents and the Proximity Effect 468 In a normal transmission line the currents are of course oppositely directed and the picture is different: F i g P . 1 3 Now the B field is larger on the right side of wire #1 so the eddy current is larger there causing the total current density J z to be larger there compared to the left side. The currents are now crowded on the side of each wire facing the other wire ( ω > 0). The Lorentz force now causes the two wires to repel each other, reversing the argument given above ( ω ≥ 0). There is still extra Ohmic loss compared to DC since Jz is non-uniform. Closer wire spacing again results in increased asymmetry. Companies like "Monster Cables" advocate using their low-ohm expensive cables for driving audio speakers in order to offset the resistance in crease due to both proximity and skin effects. Assuming a uniform J z (zeroth order), it is not hard to compute the total magnetic field B of the two conductors at any point (x,y) in the cross-secti on plane. In the following graphs we show |B y(x,y=0)| (red) as a function of x in the y = 0 plane. The wires have radius 1/2 unit and center separation 3 units (the straight-looking red curve segments are not exactly straight) : Currents in same direction: Fig P.14 Currents in opposite direction: Fig P.15 Appendix P : Eddy Currents and the Proximity Effect 469 These graphs support the claims made above concerning where the total B field is large and small, and thus where the eddy current is large and small. Of course the graphs are approximate since J z is in fact not uniform on each wire cross section. If we plot B y(x,y=0) instead of |B y(x,y=0)|, these plots have the following form, I same direction Fig P.16 I opp direction Fig P.17 In terms of our earlier drawings, the skin and proximity effects for a transmission line can be illustrated by the following top view of Fig P.13 which shows only eddy currents : Fig P.18 P.9 Quantitative Evaluation of Eddy Currents and The Proximity Effect Our round-wi re discussion above has all been qualita tive, and no method was given for computing the actual size of the eddy currents and thus of the proxi mity effect for two parallel round wires. G. Smith presents the following intriguing graph showing the strength of the proximity effect versus conductor separation for currents in the same direction. In his case c/a = 1.0 the conductors are just touching. Appendix P : Eddy Currents and the Proximity Effect 470 Fig P.19 Presumably ω is high enough to put the conductors into the skin effect regime so all currents are surface currents of thickness δ << a. We present in Section 6.5 a quantitative treatment of the skin and proximity effects for an infinite (or properly terminated) transmission line consisting of pa rallel round wires, and the results are similar to those of the above graph with θ → π-θ. Our treatment, however, is not based on "eddy current analysis", but rather on the charge distribution on the conductor surfaces (think capacitance) and the radial charge pumping boundary condition (D.2.25) which causes inte rnal currents to be la rger where the time- changing surface charge is larger. In Section 6.5 (g) we comment on how our methods might be applied to currents flowing in the same direction. P.10 Influence of Proximity and Skin Effects on Wire Resistance Consider a small differential volum e rdθ drdz in a round wire (relative to a cylindrical coordinate system for that wire). Its cross sectional area is dA = rd θdr . This volume has resistance dR = " ρL/A" = ρ dz / dA (P.10.1) and the current through this resistor will be dI = J z(r,θ) d A . ( P . 1 0 . 2 ) The Ohmic power generated in this tiny resistor is, from P = I 2R, dP = (dI)2(dR) = [J z(r,θ)dA]2 ρ dz / dA = ρ Jz(r,θ)2 dA dz . (P.10.3) Appendix P : Eddy Currents and the Proximity Effect 471 For the coin-shaped resistor consisting of length dz of the entire round wire cross section we find then that P = ∫dP = dz ∫dA ρ Jz(r,θ)2 . ( P . 1 0 . 4 ) The total current in the wire is I = ∫dA Jz(r,θ) ( P . 1 0 . 5 ) and then from P = I2R the effective wire resistance of a cross sectional slice of wire of length dz is, R = P I2 = ρdz ∫dA [Jz(r,θ)]2 [ ∫dA Jz(r,θ) ]2 ∫dA [...] = ∫0 a dr r ∫-π π dθ [...] (P.10.6) Adding some cancelling factors of A = πa2 we get, R = R/dz = [ ρ/A ] ∫dA [Jz(r,θ)]2/A [ ∫dA Jz(r,θ)/A ]2 = Rdc <Jz2> <Jz>2 = Rdc E(Jz2) [E(Jz)]2 (P.10.7) where Rdc is the DC resistance per unit length of the wire. Our notations <> and E() mean "expected value". In elementary probability theory one writes μx = E ( X ) / / m e a n σ x2 ≡ varx = E(X2) - E(X)2 = E(X2) - μx2 // variance; σx = standard deviation so that E(X 2) [E(X)]2 = σx2+ μx2 μx2 = 1 + σx2 μx2 . ( P . 1 0 . 8 ) Thus, taking X = J z we find this result for the round wire AC resistance per unit length, R = Rdc (1 + σJz2 μJz2 ) R Rdc = 1 + σJz2 μJz2 P Pdc = [1 + σJz2 μJz2 ] // loss (P.10.9) At DC, J z is constant across the wire cross secti on so its variance is 0 and the above says R = Rdc. For any other function J z(r,θ) ≠ constant, one will have some variance σJz2 > 0 and then R > Rdc. This discussion presented for a round wire of course applies to a wire of any constant cross sectional shape. Thus, the proximity and skin effects increase the effe ctive resistance of the wires in Fig P.12 or P.13, causing an increase in the Ohmic loss. Notice that the percentage proximity/skin-effect loss is independent of the current I. Appendix Q : Properties of k( ω) and Z0(ω) 472 Appendix Q: Properties of the functions k( ω) and Z0(ω) (a) Properties of k( ω) According to Section 5.3 (a), the complex wavenumber k appearing in our standard traveling wave form ej(ωt-kz) is this: k = k( ω) = -j zy = -j (R+jωL)(G+jωC) . Fact 1: The real + imaginary decomposition of k is given by (Q.1) k = a2-c 2 - j a2+c 2 k ≡ -j (R+jωL)(G+jωC) jk = a2+c 2 + j a2-c 2 jk = zy = (R+jωL)(G+jωC) where a ≡ [(R2+ω2L2)(G2+ω2C2)]1/4 = |k| dim(a) = 1/m a > 0 c ≡ RG - ω2L C d i m ( c ) = 1 / m2 c = real, |c| < a2 Proof of Fact 1: Let q ≡ zy = (R+jω L)(G+jωC) = (RG- ω2LC) + jω(LG+RC) = c + j ω(LG+RC) = |q| ejθ => |q|2 = | (R+jωL)(G+jωC) |2 = | (R+jωL)|2|(G+jωC) |2 = (R2+ω2L2) (G2+ω2C2) = a4 |q| = a 2 cosθ = Re(q) /|q| = c/a 2 Now write s ≡ zy = (R+jωL)(G+jωC) = q = |s| eiφ |s| = |q| = a φ = θ/2 Re(s) = |s| cos φ = a cos(θ /2) = a (1 +cosθ)/2 = (a/ 2 )1+c/a2 = a2+c 2 Im(s) = |s| sin φ = a sin( θ/2) = a (1 -cosθ)/2 = (a/ 2 )1-c/a2 = a2-c 2 s = zy = a2+c 2 + j a2-c 2 k = -j zy = -js = a2-c 2 - j a2+c 2 Q E D Appendix Q : Properties of k( ω) and Z0(ω) 473 Maple verification : Fact 2: In the high frequency limit (where L ≈ Le), (Q.2) Re(k) ≈ ωLC + (RC-LG)2 8ω(LC)3/2 // LC = 1/v d and ωLC = ω/vd = βd0 Im(k) ≈ - (RC + GL) 2LC + (RC - GL)2 (RC + GL) 16ω2 (LC)5/2 ω >> R/L and ω >> G/C Proof of Fact 2 : All proofs in this appendix were first (la boriously) done by hand, then Maple was used to verify them. For example, the hand derivation of Fact 2 starts out like this: a 2 ≡ (R2+ω2L2)1/2 (G2+ω2C2)1/2 = LC (R2/L2+ω2)1/2(G2/C2+ω2)1/2 = ω 2LC ( 1 + R2/(ωL)2)1/2( 1 + G2/(ωC)2)1/2 and then the expansion (1+x)1/2 = 1 + x/2 - x2/8 is used and the derivation continues. For the result to be valid, we must therefore have R 2/(ωL)2 << 1 and G2/(ωC)2 << 1 which says ω >> R/L and ω >> G/C . Appendix Q : Properties of k( ω) and Z0(ω) 474 Rather than show all the hand-done algebra, we just let Maple do the calculation. First, for Im(k) all output details are shown: and the reader sees that these first two terms of Im (k) agree with the claim above. Maple has trouble simplifying terms in expressions when they are not isolated, hence the extra code two code lines above. For Re(k) we suppress the intermediate results to save space (using : instead of ;) in agreement with the first claim of Fact 2. Appendix Q : Properties of k( ω) and Z0(ω) 475 Fact 3: In the low frequency limit with G > 0 : (Q.3) Re(k) ≈ (ω/2) RC+GL RG Im(k) ≈ - RG - (ω2/8) (RC-LG)2 (RC)3/2 ω << R/L and ω << G/C Proof of Fact 3: Fact 4: In the low frequency limit with G = 0 , (Q.4) Re(k) ≈ RC/2 ω1/2 + C/8R ω3/2 Im(k) ≈ - RC/2 ω1/2 + C/8R ω3/2 ω << R/L Proof of Fact 4: Appendix Q : Properties of k( ω) and Z0(ω) 476 (b) Properties of Z 0(ω) Fact 5: The real + imaginary decomposition of Z 0 is given by (Q.5) Z0 = R + jωL G + jωC = z y = (a/ 2 ) [ 1+(β/α) - jσ1-(β/α) ] where a = ( R2+ω2L2 G2+ω2C2 )1/4 = |Z0| α = (R2+ω2L2) (G2+ω2C2) σ = sign(RC-LG) β ≡ RG + ω2LC Proof of Fact 5: The proof is similar to that of Fact 1. Let q = z y = R+jωL G+jωC = |q| ejθ - π < θ < π ( but see few lines below) . Then |q|2 = | R+jωL G+jωC | 2 = R2+ω2L2 G2+ω2C2 ≡ a4 => |q| = R2+ω2L2 G2+ω2C2 = a2 . Next, q = R+jωL G+jωC = R+jωL G+jωC G-jωC G-jωC = (R+jωL)(G-jωC) G2+ω2C2 = RG+jω(LG-RC) + ω2LC G2+ω2C2 so Appendix Q : Properties of k( ω) and Z0(ω) 477 Re(q) = RG + ω2LC G2+ω2C2 > 0 => - π/2 < θ < π/2 Im(q) = ω(LG-RC) G2+ω2C2 sign[Im(q)] = sign(LG-RC) = sign( θ) ≡ Σ . Then cosθ = Re(q)/ |q| = (RG + ω2LC) G2+ω2C2 G2+ω2C2 R2+ω2L2 = RG + ω2LC (G2+ω2C2)( R2+ω2L2) ≡ β/α where α ≡ (G2+ω2C2)( R2+ω2L2) and β ≡ RG + ω2LC . Now write s ≡ z y = q = |s| eiφ |s| = |q| = a = |Z 0| φ = θ/2 - π/4 < φ < π /4 Re(s) = |s| cos φ = a cos(θ /2) = a (1 +cosθ)/2 = (a/ 2 )1+β/α Im(s) = |s| sin φ = a sin( θ/2) = Σ a (1 -cosθ)/2 = Σ (a/ 2 )1-β/α giving the result s = (a/ 2 ) [ 1+β/α + jΣ 1-β/α ] where a = (R2+ω2L2 G2+ω2C2 )1/4 = |Z0| α = (G2+ω2C2)( R2+ω2L2) Σ = sign(LG-RC) β ≡ RG + ω2LC . Letting σ = -Σ = sign(RC-LG) one gets the final form, s = (a/ 2 ) [ 1+β/α - jσ 1-β/α ] where a = (R2+ω2L2 G2+ω2C2 )1/4 = |Z0| α = (G2+ω2C2)( R2+ω2L2) σ = sign(RC-LG) β ≡ RG + ω2LC . Our Maple verification of this result is a bit ugly so we omit the code. An alternate geometric derivation giving the same results begins as follows: Z 0 = R +zL G +zC = L C z + (R/L) z + (G/C) = L C z + r z + g where r = (R/L) and g = (G/C) . Appendix Q : Properties of k( ω) and Z0(ω) 478 Fig Q.1 The drawing shows the complex z plane for the function Z 0(z) = L C z + r z + g in the particular case that r > g, where the z-plane has a branch cut from -r to -g. The z values of interest are only those on the positive imaginary axis where z = j ω. Reader Exercise: Finish this derivation and obtain the results shown in (Q.5). Hint : cos(β -α) = (rg+ω2)/(AB) and sin( β-α) = ω (r-g)/(AB) where A = ω2+r2 and B = ω2+g2 . Fact 6: In the high frequency limit, (Q.6) Re(Z 0) ≈ L/C + (1/8 ω2) (RC+3GL)(RC-GL) / (C5/2L3/2) Im(Z 0) ≈ - (1/2ω)(RC-GL) / (L1/2C3/2) ω >> R/L and ω >> G/C Proof of Fact 6 : We first enter into Maple the basic expressions, Appendix Q : Properties of k( ω) and Z0(ω) 479 We then request an expansion of Re(Z 0) for large ω: The first term is the expected L/C . The 1/ω2 term can be simplified as follows: and then the result is Re(Z 0) = L/C + (1/8 ω2) (RC+3GL)(RC-GL) / (C5/2L3/2) + order(1/ ω4) . For the imaginary part we do a similar set of steps: Appendix Q : Properties of k( ω) and Z0(ω) 480 with the result Im(Z 0) = - (1/2ω)(RC-GL) / (L1/2C3/2). Fact 7: In the low frequency limit with G > 0, (Q.7) Re(Z 0) ≈ R/G - (ω2/8) (3RC+GL)(RC-GL) / (C5/2L3/2) Im(Z 0) ≈ - (ω/2) (RC-GL) / ( R1/2G3/2) ω << R/L and ω << G/C Proof of Fact 7 : The Maple expressions are entered as in Fact 6, and then: so that Re(Z 0) = R/G - (ω2/8) (3RC+GL)(RC-GL) / (C5/2L3/2) + order( ω4) Im(Z 0) = - (ω/2) (RC-GL) / ( R1/2G3/2) + order( ω3) Fact 8: In the low frequency limit with G = 0, both components diverge as 1/ ω : (Q.8) Re(Z 0) ≈ (1/ 2 )R/C 1/ω Im(Z 0) ≈ - (1/ 2 )R/C 1/ω Proof of Fact 8 : We first enter the general forms, setting G = 0, Appendix Q : Properties of k( ω) and Z0(ω) 481 The expansions are then, with the final results, Re(Z 0) = (1/ 2 )R/C 1 ω Im(Z 0) = - (1/ 2 )R/C 1 ω . These results are obvious without using Maple, Z 0 =R jωC =R ωC (j)-1/2 = R ωC e-jπ/4 = R ωC [ 1- j ]/ 2 , but the expansions provide more terms if needed. Appendix R : Belden 8281 Coaxial Cable 482 Appendix R: Belden 8281 Coaxial Cable, a Case Study Here we gradually work our way through Belden's da ta sheet for its "8281" coaxial cable, correlating the data presented there with the general theory of this document. The data sheet is available here www.belden.com/techdatas/metric/8281.pdf but we will be quoting most of it below. We note that over the decades, the parameters on this data sheet have changed slightly. At its market introduc tion more than 50 years ago, this RG-59 75 Ω coaxial cable was pretty much top of the line for general purpose RF u se and is still available today. It is often used to carry uncompressed analog video signals. Today Belden offers even better coaxial cables for use with high bandwidth digital video signals. Such cables of ten have a foam dielectric to reduce attenuation, while the 8281 cable has a solid polyethylene dielectric. Below we use various equations from our main docum ent to construct a "model" of the Belden cable, but this model is limited to "high frequency" meani ng here roughly f > 1 MHz. A more careful analysis would also produce a "low frequency" model for frequencies from DC to 1 MHz, and would then blend these two models at the boundary in some smooth manner. Of the four transmission line parameters R,L,G,C, only C is treated below as a constant in frequency, although L is roughly constant in our frequency range of interest. Often in textbook treatments, all four parameters are considered constants when expressions like k(ω ) and Z 0(ω) are plotted (see below). (a) Geometry of the cable We start with this data from the Belden specification, Appendix R : Belden 8281 Coaxial Cable 483 The center wire is solid copper with a specified diameter of 0.7874 mm, so a 1 = .7874/2 mm = 0.3937 mm = 393.7 μ . The diameter of the PE core is specified as 5.0202 mm so the radius is 5.0202/2 = 2.5101mm = 2510.1 μ. This core is surrounded by a tinned doubl e copper braid. One sometimes adds the radius (80 μ) of the fine braiding wire to the effective outer cable radius, but we shall add 11.5 μ to the radius since this makes C match the Belden data sheet value if εd/ε0 = 2.3000. So a 2 = 2510.1+11.5 = 2521.6 μ. We then have, a1 = 393.7 μ inner wire radius a2 = 2521.6 μ inside radius of the shield Obviously the 2.3000 number is not exact. We are just building a reasonable model here to try and replicate the Belden claimed cable parameters, and some parameters have to be tuned to get consistency. The double braid is not exactly the same as a solid cylindrical shell of copper, so things are approximate. As we go along here, the corresponding Maple code will be displayed. So far then, where all quantities are stated in the usual SI units (meters for a 1 and a2). From these radii one computes K from (4.6.3), K = 2 ln(a 2/a1) , (4.6.3) to get K = 3.7141, (b) Capacitance C Assu ming εd/ε0 = 2.3000, one uses (4.4.17) C = 4 πε d/K capacitance per unit length (4.4.17) to get Appendix R : Belden 8281 Coaxial Cable 484 The Belden data sheet quotes C = 68.901 pF/m. in agreement with our calculation. We regard C as a constant independent of ω. (c) Conductance G The conductance G of the dielectric is re lated to the capacitance according to G = ( σ d/εd) C (4.11.34) where σd is the dielectric conductivity. Recall now (3.3.4), σeff = ( σd + ωε'd tanL) (3.3.4) which gives the effective conductivity of the dielectric in terms of the DC conductivity σd , the real part of εd called ε'd and the loss tangent factor. For PE we know that σd ~ 10-15 so we neglect that term. We shall be using tanL = .0005 belo w, and since this is small, ε'd ≈ εd. Finally, to reduce symbol clutter we rename σeff to be σd so the above equation becomes σ d = εd tanL ω => ( σd/εd) = tanL ω and then G = tan L ω C = tan L 2πf C . ( R . 1 ) Although Fig 3.1 mentions tan L = .0002 for a high quality sample of polyethylene, our experience has shown that for the bulk low-cost PE product used in coaxial cables, tan L ≈ .0005. The larger loss is due to many effects including milling, aging (oxidation), water absorption ("treeing") and additives intended to reduce these loss effects. Very poor quality PE can have a loss tangent (tan L= tanδ = dissipation factor) of .0075. For more accuracy, one can de velop frequency dependent models for tan L . The corresponding Maple expressions are duly entered, where the last notation indicates that G(f) is a function of frequency f. Appendix R : Belden 8281 Coaxial Cable 485 In either the Z 0 or attenuation calculations below, the quantity G + j ωC ( = y ) always appears as a grouping, and we have determined that G + jωC = tanL ω C + jω C = (-j tan L + 1) jω C = (1 - 0 .0005j ) j ωC . (R.2) Based on this grouping, one sees that our mode l is not very sensitive to the value of tan L as long as it is a relatively small value. (d) External inductance L e Using the numbers developed so far, one computes L e from Le = μ0 4π K in (4.11.34) : which is L e = 371.41 nH/m. (e) Total DC Inductance From (C.3.10) we know that the center wi re DC internal inductance is given by Li(center) = μ0 8π = 50 nH/m DC (C.3.10) Assuming the shield has a thickn ess t, we know from (C.6.8) that Li(shield) = μ0 8π [ (4/3)(t/a 2) ] DC thin shell, valid for t << a 2 . (C.6.8) We can compute an effective shield thickn ess t by making use of its DC resistance R 2DC: R2DC = ρ/A = 1/(σA) = 1/(σ 2πa2t) => t = 1/( σ 2πa 2R2DC) ( R . 3 ) Then Li(shield) = μ0 8π [ (4/3) 1 σ 2πa22R2DC ] . ( R . 4 ) The value of R 2DC is specified as 3.6091 ohms/km, Appendix R : Belden 8281 Coaxial Cable 486 Here then are some calculations leading to a total DC inductance for the cable, The center wire contributes 50.00 nH/m to the DC in ternal inductance, while the shield contributes another 7.96 nH/m giving a total of 57.96 nH/m for the total cable internal DC i nductance. When this is added to the external inductance L e of 371.41 nH/m, the total is seen to be 429.37 nH/m. The Belden data sheet quotes 429.811 nH/m giving a small discrepancy of 1/10th of 1% compared to our calculation, We see that Belden's "nominal inductance" is the total DC inductance of the cable L e + Li . (f) High Frequency Inductance and Resistance At high ω where the skin effect dominates, one thinks of the current being restricted to a sheath of approximate thickness δ (skin depth). In Chapter 2 we showed that, for the center conductor of radius a 1, the high frequency resistance and internal inductance (per unit length) are given by, Appendix R : Belden 8281 Coaxial Cable 487 R1 = 1 σ(2πa1)δ (2.4.18) L1i = (1/ω ) R1 = 1 ωσ(2πa1)δ = 1 ωσ(2πa1)2/ωμ0σ = 1 2πa1 μ0 2σω . (2.4.19) It was noted that R 1 has the simple interpretation of being the resistance of a shell of radius a 1 and thickness δ. Since this same kind of thin skin-e ffect sheath also exists on the inner surface of the outer conductor, we shall assume that the corresponding parameters for the outer conductor are obtained by replacing a 1 by a2 in the above, so R 2 = 1 σ(2πa1)δ L2i = 1 ωσ(2πa2)δ = 1 ωσ(2πa2)2/ωμ0σ = 1 2πa2 μ0 2σω . (R.5) We have assumed that the shield and center conduc tor are made of the same metal (copper) with σ and δ. For other cables, the shield might be aluminum foil, and one would then adjust the above equations. Adding, we then arrive at these expressions for high frequency resistance and internal inductance: R = 1 σ(2π)δ [ 1 a1 + 1 a2 ] L i = R ω = 1 ωσ(2π)δ [ 1 a1 + 1 a2 ] = 1 fσ(2π)2δ [ 1 a1 + 1 a2 ] . (R.6) At 1 MHz (2.3.9) says δ = 66μ . Since the center conductor has a 1 = radius 394 μ and the shield has thickness t = 301 μ, we shall restrict our model to apply only to frequencies over 1 MHz (ballpark). For the Belden 8281 cable, the first term in [ 1 a1 + 1 a2 ] is 6.4 times larger than the second term so most of the R and L i at high frequency come from the inner conductor, not the shield. (g) The Tinning Correction A model complication is that the 160 μ diameter copper braid wires (34 gauge) of the shield are coated with tin of thickness 1.3 μ (50 micro-inches). This coating is a dded to prevent the copper shield from oxidizing. At 1 GHz (2.3.9) gives δcopper = 2.09μ. Since tin has about 6.3 times more resistance than copper, and since δ = 2/ωμσ , one finds that δtin = 5.24μ at 1 GHz. As the frequency increases, one has Appendix R : Belden 8281 Coaxial Cable 488 to somehow gradually replace the copper δ with the tin δ in the second term of the R and L i expressions above. An analytic solution to this problem can be found by applying th e Helmholtz equation [ ∇2+β2]E = 0 to a simple one-dimensional model of the tin/copper interface. We have done this and then obtained the following "phenomenological" model to handle the tinning correction: R = 1 σ(2π)δ [ 1 a1 + 1 a2 * tf ] L i = 1 fσ(2π)2δ [ 1 a1 + 1 a2 * tf ] L i = R /(2πf) tf = 1.765 + 0.8 tanh( f(GHz) - 1.9) " tinning factor" (R.7) Here is a plot of this tinning factor for f ranging from 10 KHz to 1 GHz, Fig R.1 Tinning Factor versus Frequency It was shown above that 1 a 1 is 6.4 times larger than 1 a2 , so the tinning factor correction is fairly small at frequencies below 1 GHz (our region of interest). Even at 1 GHz we have Rcorrected Runcorrected = 1 a1 + 1 a2 * tf 1 a1 + 1 a2 = 1.026 Appendix R : Belden 8281 Coaxial Cable 489 so the tinning factor increases R and L i by about 2.6% at 1 GHz, and less below 1 GHz. Although small, we shall include this tinning correction in our calculations below. Here then are the Maple entries for high-frequency R and L i, where δ = 2/(ωμσ) = 1/(πfμσ) : Since δ ~ 1/ f and L i ~ 1/(δ f) ~ 1/ f , the internal inductance L i drops off rapidly at high frequencies and is in general much smaller than L e. This is due to the fact shown in (C.6.8) that the internal inductance of an annular shell goes to zero as that shell (thickness δ) becomes thinner. Here is a plot of L i (red), Le = 371.4 nH (black), and L = L i+ Le (green) for f in the range 1 MHz to 1 GHz, Fig R.2 Thus, for our range of interest, L is dominated by L e. The corresponding plot of resistance R is the following, Appendix R : Belden 8281 Coaxial Cable 490 Fig R.3 Notice that this R is in ohms/m, wh ereas the DC resistances of the center wire and shield are stated in ohms/km, Compared to these DC resistances, resistance R is quite la rge, and of course this is due to the skin effect. (h) Characteristic Impedance Although the cable has a nominal Z 0 of 75Ω, the actual Z 0 is a slow function of frequency and can vary slightly (~1.5 Ω) from the advertised nominal value. Recall from Chapter 4 that Z0 ≡ V(z)/i(z) = z y = R + jωL G + jωC (4.11.16) which is in general complex. We enter this into Maple, Appendix R : Belden 8281 Coaxial Cable 491 where the functions R(f), L(f) and G(f) have been stated above. We then plot Re{Z 0} for f ranging from 1MHz to 10 GHz, F i g R . 4 Recall that our cable model usi ng high frequency expressions for L i and R is only valid above 1 MHz more or less. The plot shows that the cable has Z 0 = 75Ω near f = 2 MHz, but drops to 73.48 Ω at 1 GHz, and is a little larger than 75 Ω below 2 MHz. The imaginary part of Z 0 over this same frequency range is on the order of - 1 Ω : Fig R.5 A proper no-reflections termination of the cable t hus requires both a resistance on the order of 75 Ω and a small reactive component. Since the imaginary part is so small, there is little distinction between Re(Z 0) and |Z 0|. This is illustrated in the following plot, Appendix R : Belden 8281 Coaxial Cable 492 F i g R . 6 where the red ( Re(Z 0) ) and green ( |Z 0| ) curves lie right on top of each other. At large ω, we expect Z 0 to approach a limiting value of 73.42Ω , Z0 = L/C → Le/C in agreement with Fig R.4 above. At very low ω, we have instead that Z 0 = R/G . One can use R = .036 Ω/m by adding the DC resistances of the shield and center conductor. Howeve r, G is miniscule at DC since polyethylene is such a good insulator, so Z 0 is in the 3 M Ω range, Appendix R : Belden 8281 Coaxial Cable 493 Here we have assumed σd ~ 10-15 mho/m, though this could be much larger for the kind of PE that is used in Belden cables, resu lting in a somewhat smaller Z 0DC. Our model does not account for micro detail involvi ng the mesh shield and manufacturing variations, and one finds with a network analyzer (and an actua l piece of Belden 8281 cable) that there is "noise" superimposed on our idealized plot of Z 0 versus f which has an RMS value on the order of 1 ohm, see the work of Van Der Burgt. He argues that due to this "noise", it makes little sense to try to pin down a Z 0 tolerance beyond current values, a lthough cable makers still try to do it as part of their marketing specmanship wars. (i) Phase Velocity and Attenuation Recall (5.3.6) which we appl y to the voltage on a transmission line whose left end is at z = 0: V(z) = V(0) e-jkz = V(0) e-az e-jbz jk = a + jb = zy = (R+jωL)(G+jωC) a ≡ Re( zy ) = Re[ (R+jωL)(G+jωC) ] = - Im(k) // attenuation per distance of F(z) b = Im( zy ) = Im[ (R+jωL)(G+jωC) ] = Re(k) . // phase of F(z) (5.3.6) Conventional symbols for the attenuation and phase constants are α and β, but here we call them a and b. Phase Velocity One can see that, for large ω, b = Im[ (R+jωL)(G+jωC) ] = Im[ (jωL)(jω C) ] = ω Im[ -LC ] = ω LC and therefore the cable phase velocity is given by v phase = ω/Re(k) = ω /b = 1/ LC . Appendix R : Belden 8281 Coaxial Cable 494 Since L ≈ Le in our frequency range of interest, and since th e speed of light in the dielectric is determined by vd = 1/ LeC = 1/ μdεd [see (4.11.28) ], we conclude that vphase = vd = 1/ LeC = 1/μdεd and we can compute this two different ways, knowing that the result must be the same, This is in agreement with the Belden claim, The time for a phase front to move 1 meter is given by 1/v d , which is 5.059 nsec. Belden gives which is within 1/10th a 1% of our computed value. Reader Exercise: Derive an expression for group velocity v g in the presence of attenuation (k is complex). How does your result compare with the classical expression 1/v g = ∂k/∂ω or v g = ∂ω /∂k ? Using expressions of the model above, compute v g as a function of frequency. Since v g varies with f, the cable exhibits dispersion -- pulses spread out as they are attenuated. Determine the group delay for a narrow pulse to travel 1 m down the Belden cable. How does this delay compare with the phase delay noted above? What is the effect of the tinning correction on group delay? Appendix R : Belden 8281 Coaxial Cable 495 Attenuation It is traditional to express attenuation in "vo ltage decibels" define d in this manner, dB(z) = - 20 log 10(voltage attenuation over distance z) '' decibels" = - 20 log 10(e-az) = 20 az log 10(e) = [ 20 log 10(e)] az = 8.686 az (R.8) Belden provides attenuation data for z = 100 m of cable, so we just write dB = 868.6 a = 868.6 [ - Im(k) ] > 0 . (R.9) Here then is a plot of attenuation for fre quency f in the range 1 MHz to 1 GHz : Fig R.7 In order to compare this theoretical attenuation predic tion with Belden's provided data, we first evaluate our attenuation at the frequencies lis ted on the Belden data sheet, Appendix R : Belden 8281 Coaxial Cable 496 Model Calculation of Attenuation Belden's Datasheet Attenuation The following spreadsheet then compares Belden's decibel attenuati on data with our model prediction, As shown, the error goes down as one moves away fro m the lower part of the frequency range where the model is least applicable. Here is a plot of the results, Appendix R : Belden 8281 Coaxial Cable 497 F i g R . 8 We have included in the spreadsheet a model column which ignores the tinning correction (yellow triangles). This column was obtained by setting tf(f) = 1 in the Maple code. We added a point at 2 GHz for which Belden gives no data, and at that point on e sees that the tinning correction starts to become a little more visible. Tinning increases attenuation at high frequencies. At one time we measured the attenuation of 100 m of Belden 8281 cable using a network analyzer and found that the above model (w ith tinning correction) reasonably re presents the cable up to 100 GHz. References 498 References Referen ces are in alphabetical order by the last name of the (first) author. For broken links, the referenced item can usually be found by a quick web search on th e item title. [ links all verified 31 July 2014 ]. B.I. Bleaney and B. Bleaney, Electricity and Magnetism, 3rd Ed. (Oxford University Press, London, 1976). That would be Brevis Bleaney and wife Betty Isabelle. Brevis pioneered electron spin resonance independently with Russian Yevgeny Zavoisky in 19 44. This book was reissued in 2013 as a two-volume paperback set. Chapter 10 on dielectrics is the first chapter of the second volume. R.F. Eaton and C.J. Kmiec, "Electrical Losses in Co axial Cable" (Proceedings of the 57th International Wire and Cable Symposium, Nov. 2008), www.ecadigitallibrary.com/pdf/IWCS08/14_2.pdf . [GR7] I.S. Gradshteyn and I.M. Ryzhik, Table of Integrals, Series, and Products, 7th Ed. ( Academic Press, New York, 2007). Editor Dan Zwillinger has been collecting errata. A PDF version exists. H.A. Haus and J.R. Melcher, Electromagnetic Fields and Energy (Prentice-Hall, New Jersey, 1989). Though out of print and hard to get, this very detailed and practical book is alive and well on the MIT OpenCourseWare website where all chapters can be r ead and downloaded. Two of the instructors are the authors. http://ocw.mit.edu/resources/res-6-001-electromagnetic-fiel ds-and-energy-spring-2008/ C.L. Holloway and E.F. Kuester, "DC Internal Inducta nce for a Conductor of Rectangular Cross Section", IEEE Transactions on Electromagnetic Compatibility, Vol. 51, No. 2, pp. 338-344, May 2009. J.D. Jackson, Classical Electrodynamics , 3rd Ed. (John Wiley & Sons, New York, 1998). The author was fortunate to have learned his E&M from Da ve Jackson circa 1971 (green 1st edition). R.W.P. King, Electromagnetic Engineering (McGraw-Hill, New York, 1945). This is the first of twelve books that Ronold King wrote or co-authored. His last was an antenna book (his specialty) published in 2002; he died in 2006 at age 100. It happens that th e author did an "independent study" with Prof. King circa 1969, but regrettably knew so little that Prof . King could only smile and be encouraging. [TLT] R.W.P. King, Transmission Line Theory (Dover, 1965). Another of the twelve books. C. Kittel, Introduction to Solid State Physics, 4th Ed. (John Wiley & Son, New York, 1971) . P. Lorrain, D.R Corson, F. Lorrain, Electromagnetic Fields and Waves, 3rd Ed. (W.H. Freeman & Co., New York, 1988). The first and second editions (wit hout the third author) were published in 1962 and 1970. These authors have written various other books on related topics at least through 2006. P. Lucht, Bipolar Coordinates and the Two-Cylinder Capacitor (2014). This document and the one you are reading are downloadable at http://user.xmission.com/~rimrock . If not there, search on or the document title. References 499 P. Moon and D.E. Spencer, Field Theory Handbook, Including Coordinate Systems, Differential Equations and their Solutions (Springer-Verlag, Berlin, 1961). This book is not about quantum field theory or anything like that, it is about curvilinear coordinate systems, how the Laplace and Helmholtz equations appear in each system, and what the solu tions of these equations lo ok like. This husband and wife team wrote several excellent books. Long ago they were strangely involved in an accident involving a test of general relativity. P.M. Morse and H. Feshbach, Methods of Theoretical Physics ( McGraw-Hill, New York, 1953). This 2000 page 2-volume classic behemoth is simply amazing. R. Nevels and C-S Shin, "Lorenz, Lorentz, and the Gauge", IEEE Antennas and Pr opagation Magazine , Vol 43, No 3, June 2001, pp 70-71. See www.engr.mun.ca/~egill/index_files/7811_w10/ loren z_gauge.pdf and elsewhere. [NIST] F.W.J. Olver, D.W. Lozier, R.F. Boisvert and C.W. Clark, NIST Handbook of Mathematical Functions (Cambridge University Press, 2010). NIST is the U.S. National Institute of Standards and Technology which published the world-famous earlie r edition in 1964 with editors Abramowitz and Stegun, known affectionately as "A&S". The grea tly expanded 2010 edition (968 p) can be accessed online at dlmf.nist.gov which also has errata. The book ( ≥ $17) comes with a CD containing a bookmarked PDF file which of course has been boot legged onto the web. Olver died in 2013. K.E. Oughstun, "EE 141 Lecture Notes Topic 15" (School of Engineering, University of Vermont, 2012). See http://www.emba.uvm.edu/~keoughst/Lectur eNotes141/Topic_15_(Capacitance).pdf W.K.H. Panofsky and M. Phillips, Classical Electricity and Magnetism, 2nd Ed. (Addison-Wesley, Reading MA, 1962), reissued as a Dover paperback in 2005. Some of the fascinating history of Prof. Wolfgang "Pief" Panofsky appears in Dave Jackson's Jan 2009 Physics Today article "Panofsky agonistes" which can be found at http://www-theory.lbl.gov/jdj/PT_article.pdf . H. Pender and W.A. Del Mar Editors, Handbook for Electrical Engineers, 2nd Ed (John Wiley & Sons, New York, 1922). D.B. Pengra, J. Stoltenberg, R. Van Dyck, O. Vilches, "The Hall Effect" (University of Washington, Dept of Physics, 2007). http://courses.washington.edu/phys4 31/hall_effect/hall_effect.pdf . A.D. Polyanin, Handbook of Linear Partial Differential Equations for Engineers and Scientists (Taylor & Francis, CRC Press, 2001). Polyanin and his Russian friends have recently p ublished a whole bookshelf of fat and excellent handbooks. One deals with non-linear PDEs, another with integral equations. Search for him at http://www.taylorandfrancis.com/search/ and on the web. A.M. Portis, Electromagnetic Fields: Sources and Media (John Wiley & Sons, New York, 1978). D.M. Pozar, Microwave Engineering, 4th Ed . (John Wiley & Sons, New York, 2012). Chapters 2 and 3 concern transmission lines. References 500 C. Quigley, "On the Origins of Gauge Theory" (2003), www.math.toronto.edu/~colliand/ 426_03/Papers03/C_Quigley.pdf . R.K. Rajput, Power System Engineering (Laxmi Publications, 2006), see Google books. N.J. Siakavellas, "Two Simple Models for Analytic al Calculation of Eddy Currents in Thin Conducting Plates", IEEE Trans. on Magnetics, Vol 33, No. 3, May 1997, pp 2245-2257. G. Smith, "The Proximity Effect in Systems of Para llel Conductors and Electri cally Small Multiturn Loop Antennas" (Harvard University Division of Engineering and Applied Physics, Technical Report No. 642, Dec 1971) see www.dtic.mil/dtic/tr/f ulltext/u2/ 736984.pdf . This document contains a long list of references. See also the search engine at www.dtic.mil/dtic . W.R. Smythe, Static and Dynamic Electricity, 2nd Ed . ( McGraw-Hill, New York, 1950). M. Spiegel, S. Lipschutz, and J. Liu, Schaum's Outlines: Mathematical Handbook of Formulas and Tables (4th Ed.) , (McGraw-Hill, 2012). The excellent original 1968 edition by Murray Spiegel has been a dog-eared reliable friend for many years. J ohn Liu was added for the 1999 2nd Ed, and Seymour Lipschutz joined for the 2008 3rd Ed. Not to be confused with a watered-down "Easy Outline" version. This low-cost paperback is an excellent fast refe rence for well-known mathematical facts. Our equation numbers refer to the 1968 edition. I. Stakgold, Boundary Value Problems of Mathematical Physics, Volumes 1 and 2 (MacMillan, London, 1967). These are astoundingly good books, but the high level of detail (the subject is intrinsically complex) makes them hard to use in a normal "course" , which is why the author later put out a condensed single-volume version Green's Functions and Boundary Value Problems, now in a third edition. The original two volumes were reprinted with some corr ections in 2000 ( SIAM, Philadelphia). P. Lucht has a short list of errata on line. W.T. Thomson (aka Lord Kelvin), "Ether, Electricity and Ponderable Matter" , The Proceedings of the Institution of Electrical Engineers (founded 1871), Volume 18 (1889), No 77, pp 4-37. The Appendix with ber and bei begins on page 35. Google Books has an unrestricted scan of a Harvard library copy of Vol. 18 which can be downloaded in PDF format: http://books.google.com/books?id=Wy89AAAAYAAJ [RDE] M.E.V. Valkenburg and W.M. Middleton (editors), Reference Data for Engineers: Radio, Electronics, Computers and Communications, 9th Ed. (Newness/Elsevier, Boston, 2001). Some of this document exists in Google book preview form. M. J. Van Der Burgt, "Precision Video Cables Part 1: Impedance" (Belden Electronics Division, 2002). http://www.belden.com/docs/upload/Precision-Video-Cables-Part-1.pdf . Part 2 is about return loss.